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nika2105 [10]
3 years ago
5

In rhombus MNPQ, diagonals MP and NQ intersect at point K. If m_MQP = 56', then which of the

Mathematics
1 answer:
Murrr4er [49]3 years ago
3 0

the answer is 90 degrees

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What is 3/2m-m=4+1/2m
Annette [7]

Answer:

answer is 08

Step-by-step explanation:

3/2m-m=4+1/2*m

3m/2-m=4+1/2*m

3m/2-m=4+1/2m

3m/2-m=4+1m/2

3m/2-m=4+1m/2

3m/2-m=4+m/2

3m/2-m=4+m/2

2(3m/2-m)=2(4+m/2)

m=m+8

5 0
3 years ago
One kilometer is equal to 0.62 miles. Which best describes the relationship?
Alex_Xolod [135]
C. non linear this will be the answer
5 0
3 years ago
F(x)=2x^2-3x+7 evaluate f(-2)
Vilka [71]

Answer:

Step-by-step explanation:

Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?

Let  p(x)=kpx+dp  and  q(x)=kqx+dq  than

f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7  

p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7  

(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)  

So you want:

−2kpk2q=0  

and

kpkq=−1  

and

2kpd2p−3kpdq+7=0  

Now I amfraid this doesn’t work as  −2kpk2q=0  that either  kp  or  kq  is zero but than their product can’t be anything but  0  not  −1 .

Answer: there are no such linear functions.

6 0
2 years ago
A quantity, y, varies directly as x When y=10, x=6.<br> Find xwhen y= 14.
grigory [225]

The value of x when y = 14 is 8.4

<h3 />

<h3>How to solve variation ?</h3>

y, varies directly as x. Therefore,

y ∝ x

Therefore,

y = kx

where

  • k =constant of proportionality

Therefore,

10 = 6k

k = 10 / 6

k = 5 / 3

Hence, when y = 14

y = kx

14 = 5 / 3 x

cross multiply

42 = 5x

x = 42 / 5

x = 8.4

learn more on variation here: brainly.com/question/20707320

#SPJ1

7 0
2 years ago
A ball is thrown with an initial velocity of 70 km per second, at an angle  of 35° with the horizontal. Find the vertical and ho
oksian1 [2.3K]

Answer:

i. The horizontal component of the ball = 15.8

ii. The vertical component of the ball = 11.2

Step-by-step explanation:

A thrown ball is an example of a projectile, whose motion can be explained with respect to both horizontal and vertical components.

Horizontal component of the ball = U Cos θ

Vertical component of the ball = U Sin θ

where U is the initial velocity of the ball, and θ is the angle of projection.

Thus for the given question,

U = 70 km/hr (19.44 m/s)

θ = 35°

So that;

i. The horizontal component of the ball = 19.44 Cos 35°

                                               = 19.44 x 0.8192

                                               = 15.785

The horizontal component of the ball = 15.8.

ii. The vertical component of the ball = 19.44 Sin 35°

                                                = 19.44 x 0.5736

                                                = 11.151

The vertical component of the ball = 11.2

5 0
3 years ago
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