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3 years ago

8

What can you learn about a situation from looking at a graph

1 answer:

motikmotik3 years ago

4 0

You can learn the rate for something by looking at the slope of a graph. You can also look at the start-up fee or initial cost by looking at the y-intercept of the graph. Additionally, you can get an idea of whether you are making a profit or loss in a business, how far you can throw, or many other real life situations.

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When factored, the trinomials x2 + 4x − 21 and x2 + 11x + 28 have one binomial factor in common. What is this factor?

Gelneren [198K]

Answer:

x + 7

Step-by-step explanation:

x2 + 4x − 21

set equal to 0

x2 + 4x − 21 = 0

factor

(x - 3) (x + 7)

x2 + 11x + 28

set equal to 0

x2 + 11x + 28 = 0

factor

(x + 4) (x + 7)

6 0

4 years ago

△FSB∼△ZDR .

Leya [2.2K]

Answer: 22 units.

Step-by-step explanation:

Here FSB and ZDR are two triangles.

In which, FS = 38 unit, FB = 55 unit, ZD = 15.2 unit and ZR = x unit.

Also, $\triangle FSB\sim \triangle ZDR$

Thus, By the property of similarity,

$\frac{FS}{ZD} = \frac{FB}{ZR}$

⇒ $\frac{38}{15.2} = \frac{55}{x}$

⇒ $x = \frac{55\times 15.2}{38}$

⇒ $x = \frac{836}{38}$

⇒ $x = 22$

Thus, ZR = 22 unit.

3 0

3 years ago

Read 2 more answers

Find the perimeter of quadrilateral PQRS with the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3).

storchak [24]

Answer:

$P=16.53\ units$

Step-by-step explanation:

we know that

The perimeter of quadrilateral PQRS is equal to the sum of its four length sides

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

we have

the vertices P(2,4), Q(2,3), R(-2,-2), and S(-2,3)

step 1

Find the distance PQ

P(2,4), Q(2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(0)^{2}}$

$d=\sqrt{1}$

$dPQ=1\ unit$

step 2

Find the distance QR

Q(2,3), R(-2,-2)

substitute in the formula

$d=\sqrt{(-2-3)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-5)^{2}+(-4)^{2}}$

$dQR=\sqrt{41}\ units$

step 3

Find the distance RS

R(-2,-2), and S(-2,3)

substitute in the formula

$d=\sqrt{(3+2)^{2}+(-2+2)^{2}}$

$d=\sqrt{(5)^{2}+(0)^{2}}$

$dRS=5\ units$

step 4

Find the distance PS

P(2,4), S(-2,3)

substitute in the formula

$d=\sqrt{(3-4)^{2}+(-2-2)^{2}}$

$d=\sqrt{(-1)^{2}+(-4)^{2}}$

$dPS=\sqrt{17}\ units$

step 5

Find the perimeter

$P=PQ+QR+RS+PS$

substitute the values

$P=1+\sqrt{41}+5+\sqrt{17}$

$P=6+\sqrt{41}+\sqrt{17}$

$P=16.53\ units$

5 0

4 years ago

An image has point A’(-8, 4), and the pre-image has point A(-2, 1). This is an enlargement. ASAP WILL GIVE BRAINLIEST FOR BEST A

worty [1.4K]

The answer would be true

4 0

3 years ago

Read 2 more answers

Evaluate:<br> 2 - (-9) + (-18) / 3 x 2 -10

Tatiana [17]

2+9-18/6-10
11-18/-4
-7/–4=7/4

7 0

4 years ago