what is an equation of the line that passes through the points (4,-6) and (8,-4)​

Given:

Height of rectangle = 4 units

Area of rectangle up to 48 square units.

To find:

The inequality which represents all the possible lengths in units of the bases.

Solution:

Let the possible lengths in units of the bases be b.

Area of rectangle = Length × width

Here, Height of rectangle = width = 4 units. So, area of rectangle is

[tex]Area=b\times 4[/tex]

[tex]Area=4b[/tex]

Area of rectangle up to 48 square units. It means the area should be less than or equal to 48 square units.

[tex]4b\leq 48[/tex]

Divide both sides by 4.

[tex]b\leq \dfrac{48}{4}[/tex]

[tex]b\leq 12[/tex]

Length can not be negative or zero.

[tex]0<b\leq 12[/tex]

Therefore, the required inequality is [tex]0<b\leq 12[/tex].

Answer:

H) b ≤ 12

Step-by-step explanation:

48/4 = 12.

12 is the only one you can divide into 48 equally.

X 48 ÷ 192: Can't do

X 48 ÷ 44: Can't do

X 48 ÷ 52: Can't do

√ 48 ÷ 12 = 4

Answer:

2 2/9

Step-by-step explanation:

Convert any mixed numbers to fractions.

Then your initial equation becomes:

4/3÷3/5

Applying the fractions formula for division,

=4×5

3×3

=20/9

Simplifying 20/9, the answer is

=2 2/9

Answer:

[tex]\frac{20}{9}[/tex] or [tex]2\frac{2}{9}[/tex]  or [tex]2\frac{1}{3}[/tex]

Step-by-step explanation:

[tex]1\frac{1}{3} = \frac{4}{3}[/tex] and [tex]\frac{3}{5}[/tex]

to

[tex]\frac{4}{3}[/tex] divided by [tex]\frac{3}{5}[/tex]

                    ↑

flip the second number

[tex]\frac{4}{3}[/tex] divided by [tex]\frac{5}{3}[/tex]

but instead of divison it becomes multiplication,

[tex]\frac{4}{3}[/tex] × [tex]\frac{5}{3}[/tex]

=

[tex]\frac{20}{9}[/tex]

Mixed number form is:

[tex]2\frac{2}{9}[/tex]

or

Simplify:

[tex]2\frac{1}{3}[/tex]

Answer:

(8-n) + 4

4+ (8-n)

Step-by-step explanation:

I added 4 because it says "4 more" and I subtracted 8 and a variable because it says the "difference". I used a variable because the number is not specifically defined. I hope this helps!

4

Step-by-step explanation:

Answer:

Senn needs to give Ms. Lady 4.6 minutes  head start

Senn needs to give Ms. Beetle 18.4 minutes  head start

The equation for each bug are;

For Senn, the equation for the experiment is 5 × t₁ = 92 feet

For Ms. Lady, the equation for the experiment is 4 × (t₁ + 4.6) = 92 feet

For Ms. Beetle , the equation for the experiment is 4 × (t₁ + 18.4) = 92 feet

Please find attached the required graph and table of values

Step-by-step explanation:

The given parameters are;

The average crawling time of Ms. Lady = 4 feet per minute

The average crawling time of Ms. Beetle = 2.5 feet per minute

The average crawling time of Senn = 5 feet per minute

The distance to the rose bush = 92 feet

Therefore, we have;

The time, t, duration for Senn to arrive at the Rose bush is given by the following relation,

Time, t = Distance, d/(Speed, s)

Given that the speed of the bugs is equal to their average crawling time, we have

For Senn

Time, t = 92/(5 ft/Min) = 18.4 minutes

For, Ms. Lady

Time, t = 92/(4 ft/Min) = 23 minutes

For, Ms. Beetle

t = 92/(2.5 ft/Min) = 36.8 minutes

Therefore;

Senn needs to give Ms. Lady 23 - 18.4 = 4.6 minutes head start

Similarly, Senn needs to give Ms. Beetle 36.8 - 18.4 = 18.4 minutes  head start

The equation for each bug are therefore;

For Senn, the equation for the experiment is given as follows ;

5 × t₁ = 92 feet

For Ms. Lady, the equation for the experiment is given as follows;

4 × (t₁ + 4.6) = 92 feet

For Ms. Beetle , the equation for the experiment is given as follows;

4 × (t₁ + 18.4) = 92 feet.

Answer:

y=3x+6, y=5x,

Step-by-step explanation:

Answer:

√

36

=

6

lies between consecutive odd integers  

5

and  

7

.

Step-by-step explanation:

"The" square root of  

36

is itself an integer, namely  

6

, since  

6

2

=

36

. So it does not lie between consecutive integers.

It does lie between consecutive odd integers  

5

and  

7

.

Note that  

36

has another square root, namely  

−

6

, since  

−

6

satisfies:

(

−

6

)

2

=

36

Answer:

Billy has $27.51

Step-by-step explanation:

First let's set up an equation to solve this problem.

Assign a variable to Billy's amount. Let's make it x.

Then we just interpret the problem and solve:

2x - 1.30 = 53.72

     +1.30    +1.30

----------------------------

2x = 55.02

(Divide by 2 on both sides)

----------------------------

x = $27.51

2x = Billy's amount doubled

2x - 1.30 = Taking away 1.30 from double Billy's amount

53.72 = Sara's amount

-2 + n = -8(1 - 8n) + 6

Distribute the -8

-2 + n = -8 + 64n + 6

Combine like terms.

-2 + n = 64n - 2

Add 2 to both sides and subtract n from both sides.

0 = 63n

Divide both sides by 63

0 = n

Answer:

jh

Step-by-step explanation:

uirhgurhehgu5rhginiurogherughre hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

Answer: 6.72 x 10^-5. The number is an extremely small number that started out as 6.72. You would divide the number by 10, so many times to get this number. Since the number is small, the exponent is negative. The exponent tells how many times the decimal point will be moved to the left. If the number started as 6.72 and is now 0.0000672, that means that the decimal point moved 5 times to the left.

Answer: She should us a ratio of 1:5 for grapes to blueberries

Step-by-step explanation:

Blueberries are a lot smaller than a typical grape, hence you should have 5 blueberries for every grape. Hence, the 1:5 ratio.

Answer:

(C)

Step-by-step explanation:

12:3  

Answer:

r = -3

Step-by-step explanation:

- 60= - 24 + 12r

Add 24 to each side

- 60+24= - 24+24 + 12r

-36 = 12r

Divide each side by 12

-36/12 = 12r/12

-3 =r

Answer:60x+28

Step-by-step explanation:

Answer

Kendal will be 60 meters away from the library after 80 seconds and again after 100 seconds.

Step-by-step explanation:

Let t represent time in seconds. Kendal is riding at a rate of 6 meters/second, so 6t represents the distance she can travel in t seconds.

The distance to the library is 540 meters. Because Kendal is riding toward the library, subtract 6t from 540 to find the distance between Kendal and the library before she passes it:

540 − 6t.

Kendal will be 60 meters away from the library twice, once as she approaches it and again after she passes it. The absolute value of the difference 540 − 6t represents Kendal's distance from the library. To find the time after which Kendal will be 60 meters away from the library, form an absolute value equation:

|540 − 6t| = 60.

Solve an absolute value equation by writing it as two separate linear equations and solving each one:

540 − 6t = -60 540 − 6t = 60

-6t = -600 -6t = -480

t = 100 t = 80

Let t represent time in seconds. Kendal is riding at a rate of 6 meters/second, so 6t represents the distance she can travel in t seconds.

The distance to the library is 540 meters. Because Kendal is riding toward the library, subtract 6t from 540 to find the distance between Kendal and the library before she passes it:

540 − 6t.

Kendal will be 60 meters away from the library twice, once as she approaches it and again after she passes it. The absolute value of the difference 540 − 6t represents Kendal's distance from the library. To find the time after which Kendal will be 60 meters away from the library, form an absolute value equation:

|540 − 6t| = 60.

Solve an absolute value equation by writing it as two separate linear equations and solving each one:

540 − 6t = -60 540 − 6t = 60

-6t = -600 -6t = -480

t = 100 t = 80

The answer is 80

algebra is the branch of mathematics that helps in the representation of problems or situations in the form of mathematical expressions. it involves variables like x, y, z, and mathematical operations like addition, subtraction, multiplication, and division to form a meaningful mathematical expression.

Given:

Kendal is riding at a rate of 6 meters/second

The distance to the library is 540 meters.

Let x represent time in seconds

So, kendal travel = 6x

So, distance between Kendal and the library before she passes it:

=540 − 6x.

Now, Kendal will be 60 meters away from the library.

Kendal will be 60 meters away from the library

540 − 6x= 60

-6x= 60-540

-6x= -480

x= -480/-6

x= 80

Hence, the answer is 80

Learn more about algebra here:

https://brainly.com/question/24875240

#SPJ2

Domain:  

( − ∞ , ∞ ) , { x | x ∈ R }

Range:  

[ 0 , ∞ ) , { y |y ≥ 0 }

Answer:

55.87

Step-by-step explanation:

Answer:

46 messages.

Step-by-step explanation:

The top got cut off a bit, it says "Make the same/similar inequality as the other problem:"

Answer:

The answer is D). x=2

Step-by-step explanation:

Step-by-step explanation:

2x+4 +3x = 2x +2 +1

4+x= 2x+3

4-3=2x-x

1=x

x=1

Step-by-step explanation:

this is your answer..........

Answer:

t=9

Step-by-step explanation:

Move all terms that don't contain

t to the right side and solve.t=9

Answer:

t=9

Step-by-step explanation:

48=4t+12

Subtract 12 in both sides

36=4t

Divide 4 both sides

9=t

t=9

Answer:

Step-by-step explanation:

Give the expression 3.7 x + (0.5-6y)+0.8-(1.7x+1-2.3y), to get the equivalent expression we will follow thw steps;

open the parenthesis

3.7 x + 0.5-6y+0.8-1.7x+1-2.3y

collect like terms

= 3.7 x -1.7x -6y-2.3y+0.5+0.8+1

= 2.0x-8.3y+1.3+1

= 2.0x-8.3y+2.3

Hence the equivalent expression is 2.0x-8.3y+2.3

Answer:

simplify

Step-by-step explanation:

just simpify then theres an answer waiting for you to jearney on

Answer:

[tex]x=\log_3{(15+ 4\sqrt{14})}\approx3.0949\text{ or } \\x=\log_3{(15- 4\sqrt{14})}\approx-3.0949[/tex]

Step-by-step explanation:

So we have the equation:

[tex]3^{1-x}+3^{1+x}=90[/tex]

First, note that this is the same as:

[tex]3\cdot 3^{-x}+3\cdot 3^x=90[/tex]

Factor out a 3:

[tex]3(3^{-x}+3^{x})=90[/tex]

Divide both sides by 3:

[tex]3^{-x}+3^x=30[/tex]

Change the negative exponent into a fraction:

[tex]\frac{1}{3^x}+3^x=30[/tex]

Multiply both sides by 3^x:

[tex]3^x(\frac{1}{3^x}+3^x)=(30)(3^x)[/tex]

Distribute:

[tex](1)+(3^x)^2=30(3^x)[/tex]

Let u equal 3^x. So:

[tex]1+u^2=30u[/tex]

Subtract 30u from both sides:

[tex]u^2-30u+1=0[/tex]

This is a quadratic. Solve using the quadratic formula. A is 1, b is -30, and c is 1. So:

[tex]u=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

Substitute:

[tex]u=\frac{-(-30)\pm \sqrt{(-30)^2-4(1)(1)}}{2(1)}[/tex]

Simplify:

[tex]u=\frac{30\pm \sqrt{900-4}}{2}[/tex]

Subtract:

[tex]u=\frac{30\pm \sqrt{896}}{2}[/tex]

Simplify:

[tex]u=\frac{30\pm 8\sqrt{14}}{2}[/tex]

Simplify:

[tex]u=15\pm 4\sqrt{14}[/tex]

Substitute back u:

[tex]3^x=15\pm 4\sqrt{14}[/tex]

Take the log to base 3 of both sides:

[tex]\log_3{3^x}=\log_3{15\pm 4\sqrt{14}}[/tex]

The left side cancels:

[tex]x=\log_3{15\pm 4\sqrt{14}}[/tex]

So, our solutions are:

[tex]x=\log_3{(15+ 4\sqrt{14})}\approx3.0949\text{ or } \\x=\log_3{(15- 4\sqrt{14})}\approx-3.0949[/tex]

And we're done!

Step-by-step explanation:

a-2=3+6a/3

By cross multiply

3(a-2)=3+6a

3a-6=3+6a

-6-3=6a-3a

-9=3a

a=-3