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3 years ago

I nned help understanding this math problem.

Mathematics

2 answers:

elixir [45]3 years ago

6 0

Answer:

Step-by-step explanation:

From the midsegment theorem we can conclude that

QR = $\frac{1}{2}$ NP

Using QR = 3x + 2 and NP = 2x + 16, then

3x + 2 = $\frac{1}{2}$ (2x + 16) = x + 8 ( subtract x from both sides )

2x + 2 = 8 ( subtract 2 from both sides )

2x = 6 ( divide both sides by 2 )

x = 3

As a check

QR = 3x + 2 = 3(3) + 2 = 9 + 2 = 11

NP = 2x + 16 = 2(3) + 16 = 6 + 16 = 22

Thus QR is one half NP is confirmed

poizon [28]3 years ago

5 0

Answer:

Step-by-step explanation:

QR=1/2NP ⇔ NP=2QR ⇔ 2x+16=2(3x+2) ⇔ 2x+16=6x+4 ⇔ 12=4x ⇔ x=3

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Multiply. (2x-1)(3x+5)

Nookie1986 [14]

ANSWER

6x² + 7x - 5

Step-by-step explanation:

(2x-1)(3x+5)

= 6x² + 10x - 3x - 5

= 6x² + 7x - 5

3 0

3 years ago

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X2y, for x = 3 and y = 6

DENIUS [597]

Answer:

Step-by-step explanation:

6 0

2 years ago

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Can someone please explain how to do this.

Cerrena [4.2K]

Well you can graph it by plotting 3 points then drawing a line through these 3 points. The graph will be a straight line . By plotting 3 points you can be more sure if you are right, because if you make a mistake then they might not make a straight line.

Put y = 0 in the equation and find the x coordinate:-

x + 5(0) = -20

x = -20

So we have one point to plot:- (-20,0)

Putting x = 0 we get 5y = -20 so y = -4

Second point:- (0, -4)

Putting x = 5 say we get 4 + 5y = -20 so y = -25/5 = -5

so our 3rd point is (5, -5)

7 0

4 years ago

Read 2 more answers

Given cos = 7/25, where is an angle in standard position terminating in

kozerog [31]

Answer:

tanΘ = - $\frac{24}{7}$

Step-by-step explanation:

Given

cosΘ = $\frac{7}{25}$ = $\frac{adjacent}{hypotenuse}$

The right triangle has sides 7, 24, 25 ← Pythagorean triplet

with adjacent = 7 and opposite = 24

Since position is in fourth quadrant then tanΘ < 0 , thus

tanΘ = - $\frac{24}{7}$

7 0

3 years ago

For the equation given below, evaluate y′ at the point (−2,2)<br> xe^y−4y=2x−4−2e^2

jeka94

Implicit differentiation
chain rule is important here

I'll show the steps partially
$e^y+xe^yy'-4y'=2$
$xe^yy'-4y'=2-e^y$
$y'(xe^y-4)=2-e^y$
$y'=\dfrac{2-e^y}{xe^y-4}$
now evaluate for (-2,2)
x=-2 and y=2
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{2-e^2}{-2e^2-4}$
$y'=\dfrac{e^2-2}{2e^2+4}$
that's it, simplest form

3 0

4 years ago