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m_a_m_a [10]
3 years ago
15

PLEASE HELP WITH TEST

Mathematics
1 answer:
kondor19780726 [428]3 years ago
5 0
X = 11 is the answer because IHG is scaled to be 75% of FEG.

And if you end up needed Y for that angle, Y = 25
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Find the coordinates of Rif M(-4, 5) is the midpoint<br> of RS and S has coordinates (0, -10).
alexandr402 [8]

Answer: R = (-8, 20)

Step-by-step explanation:

Given that :

M is the midpoint of RS

M = (-4, 5)

S = (0, - 10)

NOTE :

midpoint, M equals

Mx = (Rx + Sx) / 2

My = (Ry + Sy) / 2

Mx = - 4, Sx = 0, My = 5, Sy = - 10

Mx = (Rx + Sx) / 2

-4 = (Rx + 0) / 2

-8 = Rx + 0

Rx = - 8

My = (Ry + Sy) / 2

5 = (Ry + (-10)) / 2

10 = Ry - 10

10 + 10 = Ry

Ry = 20

Hence, coordinates of R = - 8 and 20

R = (-8, 20)

3 0
2 years ago
Can someone help me with these to please I’m kinda stuck on them lol
White raven [17]

Step-by-step explanation:

m+3=9

m=9-3=6

The second question I guess that it is solved like

2x+5=11

2x=11-5

2x=6

X=3

Or the other qusetion 5x-8=27

5x=19

X=19/5

I don't know the machine so sorry

7 0
2 years ago
What is the value of the 35th term in the sequence -15, -11, -7, ...? 121 125 151 374
Andreas93 [3]

Answer:

121

Step-by-step explanation:

This is an arithmetic sequence.

First term  a = -15

Common difference [d] = second term - first term

                                  = -11 - [-15] = -11 + 15 = 4

nth term = a + (n-1) * d

35th term = -15 + (35 - 1) * 4

                = -15 + 34 * 4

                = - 15 + 136

                = 121

8 0
3 years ago
Read 2 more answers
Select all the pairs that are<br> equivalent. Someone please help me. The question is number 2
ycow [4]

the answer is in the picture

8 0
3 years ago
Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 =
dexar [7]
Given f(x,\ y)=x^2-y^2-2 subject to the constraint x^2+y^2=16

Let g(x,\ y)=x^2+y^2.

The gradient vectors of f and g are:

\nabla f(x,\ y)=\langle2x,-2y\rangle and \nabla g(x,\ y)=\langle2x,2y\rangle

By Lagrange's theorem, there is a number \lambda, such that

\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle

\lambda=\pm1

It can be seen that f(x,\ y)=x^2-y^2-2 has local extreme values at the given region.
6 0
3 years ago
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