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3 years ago

What value from the set 10,11,12,13 that makes the equation 2(x-5)=12 true

Mathematics

1 answer:

Bas_tet [7]3 years ago

5 0

Answer:

Step-by-step explanation:

first (x-5) you would substitute x for 11 then subtract 5. next you would multiple by 2 and your answer would be 12

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Drayton made this model to find the product of a 3 digit number and a 1 digit number

Paha777 [63]

400x6=2400 This is how a find a product

3 0

3 years ago

Complete the equation of the line through (4,-8) and (8,5).

Alisiya [41]

Slope = (5+8)/(8-4) = 13/4

y = mx + b
b= y - mx
b = -8 - (13/4) (4)
b = -8 -13
b = -21

equation
y = 13/4x - 21

5 0

4 years ago

Fran need to move $9 million out of the country. The dimensions of a US bill is 2.61 inches by 6.14 inches by 0.0043 inches.

Mice21 [21]

Answer:

1728 in³ in a cubic ft
3.6 cubic feet (rounded to tenths)
2 suitcases

Step-by-step explanation:

For part 2, you have rounded to hundredths. The question asks for the answer rounded to tenths.

<u>Part 3</u>

(3.6 ft³)×(1 suitcase)/(2.9 ft³) = 3.6/2.9 suitcases ≈ 1.24 suitcases

Suitcases only come in whole numbers. 1 is too few, so 2 suitcases are needed.

5 0

3 years ago

What is 40+56 in GCF

KengaRu [80]

The GCF stands for greatest common factor. To represent a sum by its GCF we need to use the distributive property and we need to first find the GCF of the numbers. Let's break each number by its factors:

$\begin{gathered} 40=2\cdot2\cdot2\cdot5 \\ 56=2\cdot2\cdot2\cdot7 \end{gathered}$

We now multiply the numbers that appear on both.

$\text{GCF}=2\cdot2\cdot2=8$

We now apply the distributive property:

$8\cdot(5+7)$

3 0

2 years ago

Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i

Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

$a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333$

So the correct answer is C, the area between the two functions is 4/3 units.

3 0

3 years ago