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3 years ago

Which statement about 6x^2+7x-10 is true?

Mathematics

1 answer:

Anton [14]3 years ago

4 0

Answer:

<u>One of the factors is (x+ 2) </u>

Step-by-step explanation:

$6 {x}^{2} + 7x - 10 \\ = (6x - 5)(x + 2)$

another factor is (6x - 5)

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pickupchik [31]

I think it’s c I had this question lol

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3 years ago

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Which of the following is equal to 4 and 1 over 2 divided by 2 and 2 over 3?

Sphinxa [80]

4 and 1 over 2 times 2 and 3 over 2

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3 years ago

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Is it possible for two numbers to have a difference of 8 and a sum of 1?

topjm [15]

Answer: yes , it is possible

Step-by-step explanation:

Let the first number be x and the second number be y .Then , the sum of the two numbers will be x + y , and their difference will be x - y.

Combining the two , we have :

x + y = 1 ............................... equation 1

x - y = 8 ................................. equation 2

solving the system of linear equation by substitution method. From equation 1 , make x the subject of the formula ,

x = 1 - y ...................... equation 3

Substitute equation 3 into equation 2 ,

1 - y - y = 8

1 - 2y = 8

add 2y to both sides

1 = 8 + 2y

subtract 8 from both sides

1 - 8 = 2y

- 7 = 2y

divide through by 2

y = $\frac{-7}{2}$

y = - 3.5

substitute y = -3.5 into equation 3 to find the value of x , we have

x = 1 - y

x = 1 - ( - 3.5 )

x = 1 + 3.5

x = 4.5

Let us check :

x + y will be

4.5 + (-3.5) = 1

Also ,

x - y will be

4.5 - (-3.5)

⇒ 4.5 + 3.5 = 8

7 0

3 years ago

Determine the volume of the composite figure

madreJ [45]

1.8903
because if you do the math its easy

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3 years ago

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

4 years ago