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2 years ago

Which ordered pair is a solution to the equation below?

Mathematics

1 answer:

Nadusha1986 [10]2 years ago

4 0

3
x
−
2
y
=
10

−
3
x
+
3
x
−
2
y
=
−
3
x
+
10

0
−
2
y
=
−
3
x
+
10

−
2
y
=
−
3
x
+
10

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how do I solve 7v-(2v-16)=5(v+4) please explain if it is a one solution, no solution or infinite solution

Sergio039 [100]

Do the indicated mult. first:

7v - 2v + 16 = 5v + 20
5v + = 5v +20. This is never true. NO solution.

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3 years ago

To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

Hence final answer is -0.47.

7 0

3 years ago

I need someone to explain what I did wrong

egoroff_w [7]

Answer:

g(f(4)) = 9

Step-by-step explanation:

Fill in gf(x) for where x is in g(x): (x-7)²

Then, fill in 4 for where x is: (4-7)²

Square it: (-3)²

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2 years ago

SOLVE x^4 − 15x^2 − 16 = 0 What are the roots of the equation?

kakasveta [241]

Answer:

roots : 4, -4, i, -i

Step-by-step explanation:

This gets a bit tricky.

We have to substitude x^2 as u in this problem.

Now to rewrite x^4 − 15x^2 − 16 = 0 with u, we get

u^2 - 15u - 16 = 0

( u - 16) (u + 1)

U = 16

U = -1

This is not the end of the problem.

Now we have to substitute x^2 back to u.

x^2 = 16 --> we get the roots 4 and -4

x^2 = -1 --> we get the roots i and -i

tadah!

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