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3 years ago

Answer plz this is a test

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer:

-6 2/3

Step-by-step explanation:

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For what values of m does the graph of y = 3x2 + 7x + m have two x-intercepts

Irina18 [472]

we are given

$y=3x^2+7x+m$

To find x-intercept means we have to find zeros

and for finding zeros , we will use quadratic formula

and we have it has two x-intercepts

so, it's discriminant must be greater than 0

so, we will find discriminant

$D=\sqrt{b^2-4ac}$

now, we can compare with

$y=ax^2+bx+c$

$y=3x^2+7x+m$

and then we can find a , b and c

$a=3,b=7,c=m$

now, we can find D

$D=\sqrt{7^2-4*3*m}$

$D=\sqrt{49-12m}$

It has two x-intercepts

so,

$D=\sqrt{49-12m}>0$

now, we can solve for m

$49-12m>0$

$12m$

$m$ ............Answer

8 0

3 years ago

Read 2 more answers

9x^2-6x+8=0 what is the fifth step in solving this equation

Simora [160]

Answer:

write the equation as a binomial

Step-by-step explanation:

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3 years ago

Q # 4 Simplify the product

timama [110]

For this case we have the following product:
$(x-4) (x + 3)
$
We must use the distributive property correctly to solve the problem.
We have then:
$x ^ 2 + 3x - 4x - 12
$
Then, we must add similar terms.
We have then:
$x ^ 2 - x - 12
$
Answer:
The final product is given by:
$x ^ 2 - x - 12
$
option 2

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3 years ago

A dinner plate has a diameter of about 9 inches what is the circumference and area?

lukranit [14]

Circumference~(pi)(diameter)~28.26
Area~ (pi)(radius)squared~63.59
I believe these are right

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3 years ago

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Whats the intrgral of <img src="https://tex.z-dn.net/?f=%20%5Cint%20%20%5Cfrac%7Bx%5E2%2Bx-3%7D%7B%28x%5E3%2Bx%5E2-4x-4%29%5E2%7

rosijanka [135]

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$

5 0

3 years ago