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2 years ago

I need help with this question!!!

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

4 0

Its the first one I believe: 5/2 and 25/4

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5. If f(x) = -x2 - 7x +4 and g(x) = -5x - 3, what is the value of f(-2)-g(6)

Alex787 [66]

9514 1404 393

Answer:

Step-by-step explanation:

Evaluate the functions and find the required difference.

f(-2) = -(-2)^2 -7(-2) +4 = -4 +14 +4 = 14

g(6) = -5(6) -3 = -33

Then the difference is ...

f(-2) -g(6) = 14 -(-33) = 14 +33 = 47

5 0

2 years ago

find two numbers that have square roots between 7 and 8. one number should have a square root closer to 7 and the other should h

Bad White [126]

Step-by-step explanation:

7 = √49

8 = √64

So any positive integer between 50 to 63, both inclusive, have a square root between 7 and 8.

8 0

3 years ago

Please help me solve for x

andreev551 [17]

Answer:

x=9

Step-by-step explanation:

Corresponding angles are equal

7x - 3 = 60

7x. = 63

÷7

x = 9

Check:

7x9 - 3 = 60

Hope this helps!

5 0

2 years ago

PLEASE PELASE PLEASE HELP ME

V125BC [204]

Answer:

In ∆ABC AND ∆DEF

ABC=DEF...........each 90°

SIDE AB =SIDE ED...........given

SIDE BC =SIDE EF............B-F-C and E-C-F

∆ABC =∆DEF....................by SAS test

3 0

2 years ago

Which of the following functions are solutions of the differential equation y'' + y = 3 sin x? (select all that apply)

DiKsa [7]

Answer:

Step-by-step explanation:

We must compute the derivatives and check if the equation is satisfied.

A. $y'=(3\sin x)'=3\cos x$ . Differentiate again to get $y''=(3\cos x)'=-3\sin x$ , then $y''+y=-3\sin x+3\sin x=0\neq 3\sin x$ so this choice of y doesn't solve the equation.

B. $y'=3\sin x+3x\cos x-5\cosx +5x\sin x=(5x+3)\sin x+(3x-5)\cos x$ and $y''=5\sin x+(5x+3)\cos x+3\cos x-(3x-5)\sin x=(10-3x)\sin x+(5x+6)\cos x$ , then $y''+y=10\sin x+6\cos x\neq 3\sin x$ so y is not a solution

C. $y'=\frac{-3}{2}\cos x+\frac{3}{2}x\sin x$ hence $y''=\frac{3}{2}\sin x+\frac{3}{2}\sin x+\frac{3}{2}x\cos x=3\sin x+\frac{3}{2}x\cos x$ . Then $y''+y=3\sin x+\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\sin x$ so y is a solution.

D. $y'=-3\sin x$ and $y''=-3\cos x$ , then $y''+y=0$ thus y isn't a solution

E. $y'=\frac{3}{2}\sin x+\frac{3}{2}x\cos x$ hence $y''=\frac{3}{2}\cos x+\frac{3}{2}\cos x-\frac{3}{2}x\sin x=3\cos x-\frac{3}{2}x\cos x$ . Then $y''+y=3\cos x-\frac{3}{2}x\cos x-\frac{3}{2}x\cos x=3\cos x\neq 3\sin x$ then y is not a solution.

3 0

3 years ago