solve the log equation

Answer:

[tex]-37\frac{3}{4}[/tex]

Step-by-step explanation:

0.75 is equivalent to three quarters, or (3/4).

3/4 would be the fraction. -37 would be the whole number.

[tex]-37.75 = \boxed{-37\frac{3}{4}}[/tex]

Hope this helps.

Answer:

- 37 3/4

Step-by-step explanation:

Answer:

1/3

Step-by-step explanation:

-2-4 = -6

-7-11 = -18

Answer:

2+2=4................

Answer:

you are wrong because 2+2=4

Answer:

Rational

Step-by-step explanation:

It isnt whole because its a decimal,

It isnt an Integer because its a decimal

It isnt natural because its a decimal

Meaning it can only be rational

Answer:

7 hours

Step-by-step explanation:

Answer:

she could park there for 7 hours

Step-by-step explanation:

13-4=9 9/1.50=6

Answer:

[tex]-2^{-2} = \frac{1}{-2^{2} } = \frac{1}{4} = 0.25[/tex]

[tex]-2^{-1} = \frac{1}{-2 } = -0.5[/tex]

[tex]-2^{0} = 1[/tex]

[tex]-2^{1} = -2[/tex]

[tex]-2^{2} = 4[/tex]

Step-by-step explanation:

Answer:

86

Step-by-step explanation:

So the angle your trying to find has another angle 4 the triangle is a right triangle so 90-4=86

Hopes This Helps :)

Step-by-step explanation:

let's think this through.

train a goes from A to B in 6 hours. that means with a speed of 1/6 / hour.

train b goes from B to A in 3 hours, so it is twice as fast as train a = 2/6 / hour.

when train b leaves B (at 7pm), train a was already traveling for 2 hours (1/3 of the whole trip) leaving it with 4 hours to go (2/3 if the distance).

that means that at that point now both trains are moving against each other with a relative speed of 3 times the

speed of a (the original speed of a plus the double speed of b).

this is the same as one train standing, and the other going the whole distance with 3 times the speed of a.

the whole distance is 2/3 of AB.

the speed is 3/6 / hour = 1/2 / hour.

so, a single train with that speed would cover the total distance AB in 2 hours. or half of the distance in 1 hour.

the question now, how long for 2/3 of AB.

the distances relate by a factor :

1/2 × f = 2/3

f = 2/3 / 1/2 = 2/3 × 2/1 = 4/3

now we need to multiply also the time in the distance/time speed ratio by this factor.

therefore, 2/3 of the total distance is done in 1×4/3 = 4/3 of an hour.

that means both trains meet after 4/3 of an hour after 7pm.

that is 7pm plus 1 hour and 20 minutes giving us 8:20pm.

[tex]\rule{2000pt}{1000000pt}[/tex]

Answer:

y=5/4x-5

Step-by-step explanation:

y-y1=m(x-x1)

y-0=5/4(x-4)

y=5/4(x-4)

y=5/4x-20/4

y=5/4x-5

This value is approximate.

==========================================================

Explanation:

Let's compute the z score for x = 40

z = (x-mu)/sigma

z = (40-47)/7

z = -1

We're exactly one standard deviation below the mean.

Repeat these steps for x = 68

z = (x-mu)/sigma

z = (68-47)/7

z = 3

This score is exactly three standard deviations above the mean.

Now refer to the Empirical Rule chart below. We'll add up the percentages that are between z = -1 and z = 3. This consists of the two pink regions (each 34%), the right blue region (13.5%) and the right green region (2.35%). These percentages are approximate.

34+34+13.5+2.35 = 83.85

Roughly 83.85% of the one-mile roadways have between 40 and 68 potholes.

Answer:

132

Step-by-step explanation:

f(1) = 9

f(n) = -4f(n-1) + 4

Let n = 2

f(2) = -4f(2-1) + 4 = -4 f(1) +4 = -4(9) +4 = -36+4 = -32

Let n = 3

f(3) = -4f(2-1) + 4 =-4f(2)+4 =  -4 (-32) +4 = 128+4=132

Answer:

im not low on points but thank you <3.

Step-by-step explanation:

Answer:

:) :) :) :) :) :) :) :) :) :)

Answer:

the last one

Step-by-step explanation:

because those are oppiste poles

Answer:

OD 60 in by 45 in

Step-by-step explanation:

Answer:

see the attachment

a) The highest common factor and the least common multiple are 12 and 48, respectively.

b) The highest common factor and the least common multiple are 12 and 72, respectively.

c) The highest common factor and the least common multiple are 8 and 160, respectively.

d) The highest common factor and the least common multiple are 12 and 48, respectively.

e) The highest common factor and the least common multiple are 14 and 840, respectively.

In this question we must determine the highest common factor and the least common multiple of two integers.

The highest common factor consists in the maximum product of common prime numbers, of the two numbers that divides it, and the least common multiple consists in the product of prime numbers, common and not, of the two numbers, that creates a product that it is greater or equal to the greater integer.

Now, we proceed to determine the numbers for each case:

First, we factorize each integer:

[tex]12 = 2^{2}\times 3[/tex], [tex]48 = 2^{4}\times 3[/tex]

Then the highest common factor and the least common multiple are, respectively:

[tex]LCM = 2^{4}\times 3 = 48[/tex], [tex]HCF = 2^{2}\times 3 = 12[/tex]

The highest common factor and the least common multiple are 12 and 48, respectively. [tex]\blacksquare[/tex]

First, we factorize each integer:

[tex]24 = 2^{3}\times 3[/tex], [tex]36 = 2^{2}\times 3^{2}[/tex]

Then, the highest common factor and the least common multiple are, respectively:

[tex]LCM = 2^{3}\times 3^{2} = 72[/tex], [tex]HCF = 2^{2}\times 3 = 12[/tex]

The highest common factor and the least common multiple are 12 and 72, respectively. [tex]\blacksquare[/tex]

First, we factorize each integer:

[tex]32 = 2^{5}[/tex], [tex]40 = 2^{3}\times 5[/tex]

Then, the highest common factor and the least common multiple are, respectively:

[tex]LCM = 2^{5}\times 5 = 160[/tex], [tex]HCF = 2^{3} = 8[/tex]

The highest common factor and the least common multiple are 8 and 160, respectively. [tex]\blacksquare[/tex]

First, we factorize each integer:

[tex]24 = 2^{3}\times 3[/tex], [tex]48 = 2^{4}\times 3[/tex], [tex]60 = 2^{2}\times 3 \times 5[/tex]

Then, the highest common factor and the least common multiple are, respectively:

[tex]LCM = 2^{4}\times 3 \times 5 = 48[/tex], [tex]HCF = 2^{2}\times 3 = 12[/tex]

The highest common factor and the least common multiple are 12 and 48, respectively. [tex]\blacksquare[/tex]

First, we factorize each integer:

[tex]42 = 2\times 3 \times 7[/tex], [tex]56 = 2^{3}\times 7[/tex], [tex]70 = 2\times 5\times 7[/tex]

Then, the highest common factor and the least common multiple are, respectively:

[tex]LCM = 2^{3} \times 3 \times 5 \times 7 = 840[/tex], [tex]HCF = 2\times 7 = 14[/tex]

The highest common factor and the least common multiple are 14 and 840, respectively. [tex]\blacksquare[/tex]

To learn more on prime numbers, we kindly invite to check this verified question: https://brainly.com/question/4184435

Answer:

Well your Hypotenuse is 10 and your other angle is 8 so it is most likely 6 because you see a change in 2 each time and it is about 2 units smaller because all right triangles follow a pattern

Step-by-step explanation:

if it was too go even higher it would go to 12,14,16 etcetera

Answer:

4.546, rounded

Step-by-step explanation:

pls mark brainliest!

Answer:

y = | x - 3 | is graphed on the attachment below ↓

Step-by-step explanation:

If you ever need an equation graphed, here's a useful site called Desmos that can do that really well and for free.

Answer:

Answer:

D) The exponential function decays at three-fourths the rate of the quadratic function.

Step-by-step explanation:

Answer:

huh

Step-by-step explanation:

The distance across a circle, going through the center is called Diameter.

[tex]x = 23[/tex]

Step-by-step explanation:

First, we need to write the equations in their standard form, i.e., all the variables and their coefficients are to be placed on the left hand side and the plain numbers on the right hand side. The first equation is already in its standard form:

[tex]2x + 5y = 35\;\;\;\;\;(1)[/tex]

but we need to rewrite the 2nd one. So the standard form of the 2nd equation can be written as

[tex]x + 15y = -10\;\;\;\;\;(2)[/tex]

Now that the equations are in their standard forms, the next step that we need to do is to eliminate one of the variables, namely the y. To do that, we need to multiply either one of the equations so that when we add them together, the y-variable gets eliminated. Note if we multiply Eqn(1) by -3, we'll get -15y on one equation and a +15y on the other equation so that when they are added together, they cancel out. So if we do that, we get

[tex]-6x - 15y = -105\;\;\;\;\;(3)[/tex]

Adding Eqn(3) to Eqn(2), we get

[tex]-5x = -115[/tex]

or

[tex]x = \dfrac{-115}{-5} = 23[/tex]

Answer:

Step-by-step explanation:

Find c if ∠PQR = 90°?

I will ASSUME you mean point P is at (3. c)

slope of PQ is (2 - c) / (9 - 3) = (2 - c) / 6

slope of QR is (11 - 2) / (3c - 9) = 9 / (3c - 9)

perpendicular lines have negative reciprocal slopes.

(2 - c) / 6 = -1(3c - 9)/9

9(2 - c) = -6(3c - 9)

18 - 9c = -18c + 54

       9c = 36

         c = 4

Answer:

6.78

Step-by-step explanation:

Answer:

6.78 pounds

Step-by-step explanation:

Add: 3.18+1.35+2.25=6.78

(If you wanted the cost for the whole thing then you need to include the price for each of the pounds)

Answer:

y = m x + b    standard form of straight line

y = 2 x + 5

The slope is m and = 2

the y intercept is 5  because when x = o y = 5

if y = 0 then x = -2/5 and would be the x intercept

x is the independent variable and y is the dependent variable

I am not familiar with your definitions for domain and range but

both x and y can vary from -∝ to = ∝

Answer:

25

Step-by-step explanation:

Let,

Rhombus = ABCD

Diagonal = AC and BD

Mark its centre as O

Now,

AC = 2AO

AO = AC/2

AO = 48/2

AO = 24 cm

Also,

BD = 2BO

BO = BD/2

BO = 14/2

BO = 7 cm²

Now,

In ∆AOB

AB² = AO² + BO²

Here, AB is the side

AB² = (24)² + (7)²

AB² = 576 + 49

AB² = 625

AB = √(625)

AB = 25

Answer:

it depends how much the god snores within a certain time period