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2 years ago

Please help! I need help with this multiple choice question, thank you.

Mathematics

1 answer:

coldgirl [10]2 years ago

7 0

Answer:

A. SAS B. SSS C. AAS D. ASA

Step-by-step explanation:

A: Line segments AE and EB are congruent and line segments CE and ED are also congruent. The Angles in between line segments AE and EB and line segments CE and ED are congruent to each other

B: Line segments QT and TS are congruent. Line segments QR and RS are congruent. Both triangles share line segment TR

C: Angles A and B are congruent. Angles D1 and D2 are congruent. Line segment CD is shared between both triangles.

D: Angles C1 and C2 are congruent. Angles B1 and B2 are congruent. Line segment CB is shared between the two triangles.

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Slope intercept of 5x+2y= -2

Marianna [84]

Answer:

Step-by-step explanation:

5x + 2y = - 2

2y = - 5x - 2

y = $-\frac{5}{2}$ x - 1

5 0

3 years ago

Please help, this is for 8th grade math.

expeople1 [14]

Answer:

(2,-2)

Step-by-step explanation:

The first step is to substitute y in the first equation with the right side of the second equation because they both equal to y.

-3x+4 = 4x-10

-3x-4x+4-4 = 4x-4x-10-4

-7x = -14

-7x/-7 = -14/-7

x = 2

Plugin 2 for x for either equation to solve for y, I'll be using the second equation, but either one is fine.

y = 4(2)-10

y = 8-10

y = -2

Plugin x = 2 and y = -2 into (x,y), and you get (2,-2)

6 0

3 years ago

The perimeter of the triangle shown is 89 meters. Find the length of each side.

olga55 [171]

X=14
2x=28
4x-9=47

you solve for x after you add all the sides

3 0

1 year ago

Read 2 more answers

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

Help pls 4=36÷____________

AlladinOne [14]

Answer:

Step-by-step explanation:

36 ÷ 4 = 9

36 ÷ 9 = 4

4 0

3 years ago

Read 2 more answers