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3 years ago

Determine the x-intercept points of the function h(x) = (x – 5)(x – 6).

Mathematics

2 answers:

UNO [17]3 years ago

5 0

Answer:

(5,0),(6,0)

Step-by-step explanation:

serious [3.7K]3 years ago

5 0

<h2>Answer: (5, 0) ; (6,0)</h2>

<h3>Step-by-step explanation:</h3>

The x-intercept occurs when h(x) = 0

∴ let 0 = (x – 5)(x – 6) [using the knowledge that for the

<em> produc to be 0, one of terms </em>

⇒ x - 5 = 0 ; x - 6 = 0 <em>[solve for x in each case]</em>

⇒ x = 5 ; x = 6

<h3>∴ the points of the x-intercept are (5, 0) & (6, 0)</h3><h3 /><h3> </h3>

<em><u>Check:</u></em>

<em>If we graph the line we can see whether the intercepts are truly (5, 0) and (6,0). As seen below in the Desmos graph, it is.</em>

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HELP ME PLEASE IT WOULD BE NICE

Scorpion4ik [409]

Answer:

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Step-by-step explanation:

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3 years ago

How do you solve <br> -178=-5x+4(-4x-13)

Effectus [21]

Answer:

x = 6

Step-by-step explanation:

<em><u>Pull out like factors: </u></em>

-4x - 13 = -1 • (4x + 13)

<em><u>Equation at the end of step 2: </u></em>

-178 - (-5x + -4 • [4x + 13] ) = 0

<em><u>Pull out like factors: </u></em>

21x - 126 = 21 • (x - 6)

<em><u>Equation at the end of step 4: </u></em>

21 • (x - 6) = 0

<em><u>Solve:</u></em>

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<em><u>Add 6 to both sides of the equation:</u></em>

x = 6

5 0

3 years ago

Please answer,<br> don't just not answer for points

Goshia [24]

Answer:

I'm pretty sure the answer to number one is 12g.

Step-by-step explanation:

To combine like terms, you combine all of the same variables.

For example, if you have 5x+7y+5x, you would combine like terms to make 10x+7x.

So for your first problem, we know that the combined like term is 15g. We know that one of the numbers combined was 3g. With that knowledge, you can subtract 15g-3g and get 12g.

I hope this makes sense, and helps you answer your other questions!

5 0

3 years ago

Read 2 more answers

Use the Green's Theorem to calculate the work done by the field [ F (x, y) = -3y^5 i + 5y^2x^3 j ] to move a particle along the

lianna [129]

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\iint_R\left(\frac{\partial(5y^2x^3)}{\partial x}-\frac{\partial(-3y^5)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$\dfrac{\partial(5y^2x^3)}{\partial x}=15y^2x^2$
$\dfrac{\partial(-3y^5)}{\partial y}=-15y^2$

$=\displaystyle\int_{x=-2}^{x=2}\int_{y=-\sqrt{4-x^2}}^{y=\sqrt{4-x^2}}15y^2(x^2+1)\,\mathrm dy\,\mathrm dx$

Converting to polar coordinates, the integral is equivalent to

$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=2}15r^3\sin^2\theta(r^2\cos^2\theta+1)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle15\int_0^{2\pi}\int_0^2\left(\frac{r^5}4\sin^22\theta+r^3\sin^2\theta\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\frac{15}4\left(\int_0^2r^5\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^22\theta\,\mathrm d\theta\right)+15\left(\int_0^2r^3\,\mathrm dr\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)$
$=100\pi$

6 0

3 years ago

How is 5/9 -2/3 = 2/9 I got 3/9

kupik [55]

$\frac{5}{9}-\frac{2}{3}=\\
\frac{5}{9}-\frac{6}{9}=\\
-\frac{1}{9}
$

6 0

4 years ago