Answer:
7i
Step-by-step explanation:
We know that i = √-1
Then solving the given:
√-49 = √49×(-1) = √49 × √-1 = 7×√-1 = 7iSo the answer is:
√-49 = 7iFinding missing angles
Answer:
x=15
Step-by-step explanation:
165+x=180
180-165=15
Hope this helps!!
I NEED HELP PLEASE, THANKS! :)
Answer:
Symmetric with respect to the polar axis in agreement with the second answer listed.
Step-by-step explanation:
This is the shape of a cardioid [tex]14\,(1+cos(\theta))[/tex] it contains the function cosine of the angle so it must be symmetric with respect to the polar axis, since the cosine function is also symmetric for positive and negative values of the angle.
What is the quotient in polynomial form?
Answer:
The quotient in polynomial form= 2x + 6
Step-by-step explanation:
In order to calculate the quotient in polynomial form of the following synthetic division we would have to make the following:
According to the given data we have the following:
-1|2 8 6
Therefore, quotient in polynomial form would be calculated as follows:
-1 | 2 8 6
-2 -6
2 6 0
Therefore, quotient in polynomial form= 2x + 6
The quotient in polynomial form= 2x + 6
An object is dropped from the top of a tower with a height of 1160 feet. Neglecting air resistance, the height of the object at time t seconds is given by the
polynomial - 16t square + 1160. Find the height of the object at t = 1 second.
The height of the object at 1 second is feet.
Answer:
Height at t = 1 sec is 1144 ft
Step-by-step explanation:
Given:
Initial height of object = 1160 feet
Height of object after t seconds is given by the polynomial:
[tex]- 16t ^2+ 1160[/tex]
Let [tex]h(t)=- 16t ^2+ 1160[/tex]
Let us analyze the given equation once.
[tex]t^2[/tex] will always be positive.
and coefficient of [tex]t^2[/tex] is [tex]-16[/tex] i.e. negative value.
It means something is subtracted from 1160 ft (i.e. the initial height).
So, height will keep on decreasing with increasing value of t.
Also, given that the object is dropped from the top of a tower.
To find:
Height of object at t = 1 sec.
OR
[tex]h (1)[/tex] = ?
Solution:
Let us put t = 1 in the given equation: [tex]h(t)=- 16t ^2+ 1160[/tex]
[tex]h(1)=- 16\times 1 ^2+ 1160\\\Rightarrow h(1) = -16 + 1160\\\Rightarrow h(1) = 1144\ ft[/tex]
So, height of object at t = 1 sec is 1144 ft.
Factor this trinomial.
х^2 - 9х+ 20
А. (х- 4)(х+ 5)
В. (х-2)(х+ 10)
С. (х-2)(х- 10)
D. (x-4)(х - 5)
Answer:
D
Step-by-step explanation:
We're looking for two numbers that have a sum of -9 and a product of 40. These 2 numbers are -4 and -5 so the answer is (x - 4)(x - 5).
Answer:
(x-4)(x-5)
Step-by-step explanation:
х^2 - 9х+ 20
What 2 numbers multiply to 20 and add to -9
-4*-5 = 20
-4+-5 = -9
(x-4)(x-5)
What ray is a common side of
Which of the following lists of ordered pairs is a function?
Α. (Ο, 2), (2, 3), (ο, - 2), (4, 1)
Β. (1, 2), (1,2), 2), (3, 4)
C. 1, 5). 2, 1, 4, 9), το, 5)
D. (2, 4), (0, 2), (2, -4), (5, 3)
Answer:
the answer will be D
Step-by-step explanation:
Find the value of x.
Answer:
[tex]\huge\boxed{x=\sqrt{66}}[/tex]
Step-by-step explanation:
ΔADC and ΔABD are similar (AAA)
Therefore the cooresponging sides are in proportion:
[tex]\dfrac{AD}{AC}=\dfrac{AB}{AD}[/tex]
Substitute
[tex]AD=x;\ AC=6+5=11;\ AB=6[/tex]
[tex]\dfrac{x}{11}=\dfrac{6}{x}[/tex] cross multiply
[tex](x)(x)=(11)(6)\\\\x^2=66\to x=\sqrt{66}[/tex]
Suppose H is an ntimesn matrix. If the equation Hxequalsc is inconsistent for some c in set of real numbers R Superscript n, what can you say about the equation Hxequals0? Why?
Answer:
The answer is explained below
Step-by-step explanation:
Given that, the equation H*x = c is inconsistent for some c in R^n, we can say that the equation A*x = b has at least one solution for each b in R^n of IMT (Inverse Matrix Theorem) is not fulfilled.
Thanks to this we can say that by equivalence of theorem statement, the equation H*x = 0 will not have only the trivial solution. It will have non-trivial solutions too.
A sample of a radioactive substance has an initial mass of 58.7 . This substance follows a continuous exponential decay model and has a half-life of 5 hours. (a)Let t be the time (in hours) since the start of the experiment, and let y be the amount of the substance at time t . Write a formula relating y to t. y=__ e^__(t) . Use exact expressions to fill in the missing parts of the formula. Do not use approximations. b)How much will be present in 8 hours? Do not round any intermediate computations, and round your answer to the nearest tenth.
Answer:
a). Y = y0e^-k(t)
b) Y = 19.4 Unit mass
Step-by-step explanation:
Y = y0e^-k(t)
Where y is amount present at the time
Y0 is initial amount present at t = 0
Y0 = 58.7
Half life = 5 hours
At half life , y = 58.7/2
At half life , y = 29.35
K = decaying constant.
Let's look fithe value of k
Y = y0e^-k(t)
29.35 = 58.7e^-k(5)
29.35/58.7 = e^-k(5)
0.5 = e^-k(5)
In 0.5 = -k(5)
-0.69314718 = -k(5)
0.138629436 = k
The value present in 8 hours will be
Y = y0e^-k(t)
Y = 58.7e-0.138629436(8)
Y = 58.7e-1.109035488
Y = 58.7(0.329876978)
Y= 19.36377861
To the nearest tenth
Y = 19.4 unit of mass
Don’t know this one
Answer:
B
Step-by-step explanation:
The answer is B because in order for the square root of a number to be equal to another number, the answer squared should be the number under the square root.
B. [tex](-4)^2\neq -16[/tex].
Hope this helps.
The base of pyramid A is a rectangle with a length of 10 meters and a width of 20 meters. The base of pyramid B is a square with 10-meter sides.
The heights of the pyramids are the same.
The volume of pyramid Als
y the volume of pyramid B. If the helght of pyramid B increases to twice that of pyramid A, the
new volume of pyramid B is
the volume of pyramid A.
Answer:
a. The volume of Pyramid A is double that of Pyramid B.
b. The new volume of B is equal to the volume of A.
Step-by-step explanation:
The base of pyramid A is a rectangle with length 10 meters and width 20 meters.
The base of pyramid B is a square of side length 10 meter.
Both pyramids have the same height, h.
The volume of a pyramid is given as:
V = lwh / 3
where l = length
w = width
h = height
The volume of Pyramid A is:
V = (10 * 20 * h) / 3 = 66.7h cubic metres
The volume of Pyramid B is:
V = (10 * 10 * h) / 3 = 33.3h cubic metres
By comparing their values, the volume of Pyramid A is double that of Pyramid B.
If the height of B increases to 2h, its new volume is:
V = (10 * 10 * 2h) / 3 = 66.7h cubic metres
The new volume of B is equal to the volume of A.
Last winter Armand had StartFraction 5 Over 6 EndFraction of a row of stacked logs. At the end of the winter he had StartFraction 8 Over 15 EndFraction of the same row left. How much wood did he burn over the winter?
Answer:
3/10
Step-by-step explanation:
We have that the Armans last winter had 5/6 of a row of stacked logs and at the end of the winter he had 8/15 of the same row left, therefore:
Ambitious
First we have to do is that the denominator is the same.
in the case of 5/6 it would be 25/30
and for 8/15 it would be 16/30
Now if we can do the subtraction and it would be:
25/30 - 16/30 = 9/30 or what equals 3/10
3/10 was the amount of wood he burned in the winter
Answer:
D) 3/10 row
Step-by-step explanation:
Please answer this correctly
Answer:
2/7
Step-by-step explanation:
The numbers greater than 7 or less than 3 are 2 and 8.
2 numbers out of 7.
P(greater than 7 or less than 3) = 2/7
Answer:
2/7
Step-by-step explanation:
There are a total of 7 sample spaces also known as 2,3,4,5,6,7,8. Now we have to find a number greater than 7 and less than 3. 2 is less than 3, and 8 is greater than 7, so two numbers are selected. This would become 2/7 because out of all of the 7 outcomes, only two are selected.
Sam invests $6000 in two different accounts. The first account paid 12 %, the second account paid 7 % in interest. At the end of the first year he had earned $590 in interest. How much was in each account? $_____ at 12% $_____ at 7%
Answer:
The amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600
Step-by-step explanation:
Let x be the amount in the account at 12% interest
So, 6000-x is the amount in the second account at 7% interest
[tex]SI = \frac{P \times T \times R}{100}[/tex]
First account:[tex]SI=\frac{x \times 1 \times 12}{100}[/tex]
Second account : [tex]SI =\frac{(6000-x) \times 1 \times 7}{100}[/tex]
We are given that At the end of the first year he had earned $590 in interest.
So, [tex]\frac{x \times 1 \times 12}{100}+\frac{(6000-x) \times 1 \times 7}{100}=590\\x=3400[/tex]
So,the amount in the account at 12% interest is $3400
The amount in the second account at 7% interest =6000-x=6000-3400=2600
Hence the amount in the account at 12% interest is $3400 and the amount in the second account at 7% interest is $2600
In a random sample of 400 registered voters, 120 indicated they plan to vote for Candidate A. Determine a 95% confide
Answer:
Step-by-step explanation:
Using the formula
p +/- (z* √[p(1-p) /n]
Where p is sample proportion = 120/400= 0.3 z*= 1.96 (z score for 95% confidence) and n is 400.
0.3 + (1.96 √[0.3(1-0.3) / 400])
0.3 + (1.96 √[(0.3*0.7)/400
0.3 + (1.96√ (0.21/400))
0.3 + (1.96 √0.000525)
0.3 + (1.96* 0.023)
0.3 + (0.045)
= 0.345 ~ 35%
For the lower interview
0.3 - (0.045)
= 0.255 ~ 26%
Thus, a 95% confidence interval for this study is between 26% and 35%
(2m^4) • m3
Need help ASAP I have the answer but I need help showing work
Answer:
4m^11
Step-by-step explanation:
2m^4.2 you multiply powers in paranthesis
2m^8 result
2m^8.m^3
You add the exponents together because it is same base and put the 4 because 2 to the power of 2 is 4
4m^11
Answer: 4[tex]m^{11}[/tex]
Step-by-step explanation:
(2m^4)^2 * m^3
Expand
[tex]2^{2} * m^{8} *m^{3}[/tex] add the exponents since they have the same base
[tex]4m^{11}[/tex]
true or false? the circumcenter of a triangle is the center of the only circle that can be inscribed about it
Answer:
TRUE
Step-by-step explanation:
The circumcenter of a triangle is the center of the only circle that can be circumscribed about it
Answer:
False
Step-by-step explanation:
Find the median of: 1, 3, 4, 6, 2, 4, 5, 6, 2, 3, 1, 4, 0, 4, 4, 4, 8, 9, 7, 4
Answer:
4
Step-by-step explanation:
1, 3, 4, 6, 2, 4, 5, 6, 2, 3, 1, 4, 0, 4, 4, 4, 8, 9, 7, 4
Arrange the numbers from smallest to largest
0,1, 1,2,2, 3,3, 4, 4,4,4,4,4 , 4, 5, 6, 6, 7, 8, 9,
There are 20 numbers
The middle number is between 10 and 11
0,1, 1,2,2, 3,3, 4, 4,4 ,4,4,4 , 4, 5, 6, 6, 7, 8, 9,
The median is 4
Solution,
Arranging the data in ascending order:
0,1,1,2,2,3,3,4,4,4,4,4,4,4,5,6,6,7,8,9
N(total number of items)= 20
Now,
Median:
[tex] (\frac{n + 1}{2)} ) ^{th \: item} \\ = (\frac{20 + 1}{2} ) ^{th \: item} \\ = \frac{21}{2} \\ = 10.5 \: th \: \: item[/tex]
Again,
Median:
[tex] \frac{10 \: th \: item + 11 \: th \: item}{2} \\ = \frac{4 + 4}{2} \\ = \frac{8}{2} \\ = 4[/tex]
Two parallel lines are crossed by a transversal.What is the value of d?
Step-by-step explanation:
if there is any confusion then again ask me always with you
Answer:
d = 125
Step-by-step explanation:
E2020
pls mark Brainliest
Suppose you are told that the grades of the nal exam of your class follows a normal distribution. The average score is 65 and the variance is 12. What is the the probability of a random classmate getting a score more than 80?
Answer:
0% probability of a random classmate getting a score more than 80
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation(which is the square root of the variance) [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 65, \sigma = \sqrt{12} = 3.4641[/tex]
What is the the probability of a random classmate getting a score more than 80?
This is 1 subtracted by the pvalue of Z when X = 80. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{80 - 65}{3.4641}[/tex]
[tex]Z = 4.33[/tex]
[tex]Z = 4.33[/tex] has a pvalue of 1
1 - 1 = 0
0% probability of a random classmate getting a score more than 80
Suppose you are starting your own company selling chocolate covered strawberries. You decide to sell the milk chocolate covered strawberries for a profit of $2.25 $ 2.25 /box and the white chocolate covered strawberries at $2.50 $ 2.50 /box. Market tests and available resources, however, have given you the following constraints. The combined production level should not exceed 800 800 boxes per month. The demand for the white chocolate is no more than half the demand for milk chocolate strawberries. The production level for white chocolate should be less than or equal to 200 200 boxes.
Answer:
$1850 per month
Step-by-step explanation
There are two types of chocolates that can be produced milk chocolate and strawberry covered chocolate. To find the profit we make following equation,
P = $2.25 SC + $2.50 WC
where SC is strawberry chocolate and WC is White milk chocolate.
The maximum production level can be 800 boxes per month and white chocolates can not exceed the 200 boxes per month so we assume making 600 boxes of Strawberry covered chocolates and 200 boxes of white chocolates.
Profit = 2.25 * 600 + 2.50 * 200
Profit = $1850
This is the maximum profit that can be earned after making combination of two types of chocolates.
9(d − 93) = –36 d = _______
Steps to solve:
9(d - 93) = -36d
~Distribute
9d - 837 = -36d
~Subtract 9d to both sides
-837 = -45d
~Divide -45 to both sides
18.6 = d
Best of Luck!
Katie is making hair clips to sell at the craft fair. To make each hair clip, she uses 1 barrette and 1 precut ribbon. The barrettes are sold in packs of 12, and the precut ribbons are sold in packs of 9. How many packs of each item does she need to buy to make the least number of hair clips with no supplies left over?
Answer:
3 packs of barrettes and 4 packs of ribbons.
Step-by-step explanation:
All we need to do here is to find the least common multiple between 12 and 9.
We can factor both of these numbers to do so.
12: 3*4: 3*2*2
9: 3*3
We can cancel out one 3 (since it appears in both prime factorizations) and multiply what we have left to find the LCM.
2*2*3*3=36
This means that she will be making 36 clips/needs 36 of each item.
36/12=3
3 packs of barretes.
36/9=4
4 packs of ribbons.
The height of a projectile launched upward at a speed of 32 feet/ second from a height of 128 feet is given by the function h(t)=-16t^2+32t+128. How long will it take the projectile to hit the ground?
Answer:
8.51 seconds
Step-by-step explanation:
Time to reach maximum height
=( Usin90)/2g
= 32/2*9.81
= 32/19.62
= 1.63 s
Max height = U²Sin²tita/g
= 32²/9.81
= 1024/9.81
= 104.38ft
Total height=104.38 + 128= 232.38
Time to fall back to the ground using the formula
h = ut+(1/2)gt²
U= 0
H= 232.38
g = 9.81
(232.38 *2 )/9.81= t²
47.376= t²
t = 6.88 s
Total time taken to hit ground
= 6.88 + 1.63
= 8.51 seconds
Ujalakhan01! Please help me! ASAP ONLY UJALAKHAN01. What's (x-1)(x-1)?
Answer:
[tex]x^2-2x+1[/tex]
Step-by-step explanation:
=> (x-1)(x-1)
Using FOIL
=> [tex]x^2-x-x+1[/tex]
=> [tex]x^2-2x+1[/tex]
Answer:
[tex]\boxed{x^2-2x+1}[/tex]
Step-by-step explanation:
[tex](x-1)(x-1)[/tex]
Apply FOIL Method.
[tex]x(x-1)-1(x-1)[/tex]
[tex]x^2-x-x+1[/tex]
Combine like terms.
[tex]x^2-2x+1[/tex]
A delivery truck is transporting boxes of two sizes: large and small. The combined weight of a large box and a small box is 75 pounds. The truck is transporting 55 large boxes and 50 small boxes. If the truck is carrying a total of 4025 pounds in boxes, how much does each type of box weigh?
Answer:
There are 50 large boxes.
Step-by-step explanation:
First, "boxes of two sizes" means we can assign variables:
Let x = number of large boxes
y = number of small boxes
"There are 115 boxes in all" means x + y = 115 [eq1]
Now, the pounds for each kind of box is:
(pounds per box)*(number of boxes)
So,
pounds for large boxes + pounds for small boxes = 4125 pounds
"the truck is carrying a total of 4125 pounds in boxes"
(50)*(x) + (25)*(y) = 4125 [eq2]
It is important to find two equations so we can solve for two variables.
Solve for one of the variables in eq1 then replace (substitute) the expression for that variable in eq2. Let's solve for x:
x = 115 - y [from eq1]
50(115-y) + 25y = 4125 [from eq2]
5750 - 50y + 25y = 4125 [distribute]
5750 - 25y = 4125
-25y = -1625
y = 65 [divide both sides by (-25)]
There are 65 small boxes.
Put that value into either equation (now, which is easier?) to solve for x:
x = 115 - y
x = 115 - 65
x = 50
Write an equation that is 10 less than 3 times a number y, multiplied by 2 and divided by 4. (10 less than 3 times a number y is to be done first)
Answer: (3y - 10)*2÷4
Step-by-step explanation:
Because 10 less than 3 times a number, y, is done first, it is in parenthesis. The 3 is there to represent the "three times" and the -10 is there to represent the "ten less". The *2 is there to represent the "multiplied by two" and the ÷4 is there to represent the "divided by 4"
Hope it helps, and tyvm <3
Answer:
[tex]\displaystyle \frac{2(3y - 10)}{4}[/tex]
Step-by-step explanation:
10 less than 3 times y.
The variable y is multiplied by 3, 10 is subtracted from 3 × y.
The result 3y - 10 is then multiplied by 2.
2(3y - 10) is then divided by 4.
There are no solutions to the system of inequalities shown below.
y< 3x+ 5
y> 3x-1
True or false
Someone help please
Answer:
False, there are actually infinite solutions as these are parallel lines.
Step-by-step explanation:
What is the domain of the equation y=1/X+5?
Answer:
Domain is
{
x
∈
R
;
x
≠
−
5
}
Range is
{
y
∈
R
;
y
≠
0
}
Step-by-step explanation:
Explanation:
Domain: Denominator should not be
0
∴
x
+
5
≠
0
or
x
≠
−
5
Domain is any real value except
x
=
−
5
or
{
x
∈
R
;
x
≠
−
5
}
Range is any real value except
y
=
0
or
{
y
∈
R
;
y
≠
0
}
graph{1/(x+5) [-10, 10, -5, 5]}