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3 years ago

Solve for x and y please !!

Mathematics

2 answers:

LenaWriter [7]3 years ago

8 0

since both inside angles are 45, the height and the base would be the same

the height is 72 yd

so the base (y) would also be 72 yd

Y = 72

and x would equal y * sqrt(2)

so x = 72sqrt(2)

prohojiy [21]3 years ago

7 0

Isosceles triangle
y = 72 yd

x^2 = 72^2 + 72^2
x^2 = 2(72^2)
x = 72√2 yd

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What is the solution to the system of equations <br><br> 3x+2y=39<br> 5x-y=13

Vadim26 [7]

Hiya. I'm going to rewrite the second equation.

By subtracting 5x by both sides, I'll be able to have y by itself:

-y=-5x+13

I'm going to then divide both sides by -1 to get:

y=5x-13.

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Plug this back into the equation of y=5x-13

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3 years ago

Two vectors A⃗ and B⃗ are at right angles to each other. The magnitude of A⃗ is 5.00. What should be the length of B⃗ so that th

creativ13 [48]

Answer:

Length of B is 7.4833

Step-by-step explanation:

The vector sum of A and B vectors in 2D is

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And its magnitude is:

$C=\sqrt{(a_1+b_1)^2+(a_2+b_2)^2} =9$

Where

$a_1=Asinx$

$a_2=Acosx$

$b_1=Bsin(x+90)$

$b_2=Bcos(x+90)$

Using the properties of the sum of two angles in the sin and cosine:

$b_1=Bsin(x+90)=B(sinx*cos90+sin90*cosx)=Bcosx$

$b_2=Bcos(x+90)=B(cosx*cos90-sinx*sin90)=-Bsinx$

Sustituying in the magnitud of the sum

$C=\sqrt{(Asinx+Bcosx)^2+(Acosx-Bsinx)^2} =9$

$C=\sqrt{A^2sin^2x+2ABsinxcosx+B^2cos^2x+A^2cos^2x-2ABsinxcosx+B^2sin^2x} =9$

$C=\sqrt{A^2(sin^2x+cos^2x)+B^2(cos^2x+sin^2x)}$

$C=\sqrt{A^2+B^2} =9$

Solving for B

$A^2+B^2 =9^2$

$B^2 =9^2-A^2$

Sustituying the value of the magnitud of A

$B^2=81-5^2=81-25=56$

$B= 7. 4833$

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