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3 years ago

Help please ......................

Mathematics

1 answer:

prohojiy [21]3 years ago

5 0

Plug -2 in for x and -5 in for y.

A. 1 - 2(-2) + 3(-5) = -10

B. - |(-2) - (-5)| = -3

C. Sqrt((-2)^2) + cubert((-5)^3) = -7

D. 1/2(-2) + 1/3(-5) = -8/3

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Javen and Andrew

Tasya [4]

Answer:

joven has 21 and Andrew has 16

Step-by-step explanation:

37-5=32

32/2=16= andrew b/c he has 5 fewer

16+5=21=Joven b/c he has 5 more

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3 years ago

The length of a rectangle is 10 yd less than three times the width, and the area of the rectangle is 77 yd^2. Find the dimension

Vitek1552 [10]

Answer:

W=7 and L=11

Step-by-step explanation:

We have two unknowns so we must create two equations.

First the problem states that length of a rectangle is 10 yd less than three times the width so: L= 3w-10

Next we are given the area so: L X W = 77

Then solve for the variable algebraically. It is just a system of equations.

3W^2 - 10W - 77 = 0

(3W + 11)(W - 7) = 0

W = -11/3 and/or W=7

Discard the negative solution as the width of the rectangle cannot be less then 0.

So W=7

Plug that into the first equation.

3(7)-10= 11 so L=11

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3 years ago

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Step 2 because what if they didn't think it was delicious....

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3 years ago

What method did you use to get the answer that I submitted.

raketka [301]

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Step-by-step explanation:

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3 years ago

For what value of k are the roots of the quadratic equation kx (x-2)+6=0 equal?

ser-zykov [4K]

Answer:

$\boxed{\sf k=6}$

Step-by-step explanation:

$\sf kx (x-2)+6=0$

Expand brackets.

$\sf kx^2 -2kx+6=0$

This is in quadratic form.

$\sf ax^2 +bx+c=0$

Since this is for equal roots:

$\sf b^2 -4ac=0$

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$\sf (-2k)^2 -4(k)(6)=0$

$\sf 4k^2-24k=0$

$\sf 4k(k-6)=0$

$\sf 4k=0\\k=0$

$\sf k-6=0\\k=6$

Plug k as 0 to check.

$\sf \sf 0x^2 -2(0)x+6=0\\6=0$

False.

So that means k must equal 6.

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4 years ago