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Inessa05 [86]
2 years ago
13

At what point does the graph of y=9x(x is exponent) cross the y-axis?

Mathematics
1 answer:
uysha [10]2 years ago
3 0

Answer:

-<u>T</u><u>h</u><u>e</u><u> </u><u>g</u><u>r</u><u>a</u><u>p</u><u>h</u><u> </u>crosses <u>t</u><u>h</u><u>e</u><u> </u>y-axis <u>a</u><u>t</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>p</u><u>o</u><u>i</u><u>n</u><u>t</u><u> </u><u>(</u>0,0)

<u>N</u><u>o</u><u>n</u><u>e</u><u> </u><u>o</u><u>f</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>g</u><u>i</u><u>v</u><u>e</u><u>n</u><u> </u><u>o</u><u>p</u><u>t</u><u>i</u><u>o</u><u>n</u><u>s</u><u> </u><u>i</u><u>s</u><u> </u><u>c</u><u>o</u><u>r</u><u>r</u><u>e</u><u>c</u><u>t</u><u> </u>

<u>A</u><u> </u><u>g</u><u>r</u><u>a</u><u>p</u><u>h</u><u> </u><u>c</u><u>r</u><u>o</u><u>s</u><u>s</u><u>e</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u>y-axis <u>a</u><u>t</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>p</u><u>o</u><u>i</u><u>n</u><u>t</u><u> </u><u>(</u>0,y).<u>T</u><u>h</u><u>i</u><u>s</u><u> </u><u>m</u><u>e</u><u>a</u><u>n</u><u>s</u><u> </u><u>t</u><u>h</u><u>a</u><u>t</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>g</u><u>r</u><u>a</u><u>p</u><u>h</u><u> </u><u>w</u><u>i</u><u>l</u><u>l</u><u> </u><u>c</u><u>r</u><u>o</u><u>s</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u>y-axis <u>w</u><u>h</u><u>e</u><u>n</u><u> </u>x-=0

To find the point where the graph crosses the y-axis,substitute x=0 into the given equation y-9x

substitute x=0 into the equation y-9x to find the corresponding value of y

That is,

y=9(0)

y=0

therefore the graph crosses the y-axis at the point (0,0)

Step-by-step explanation:

hope it hlps

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Chase's birthday party costs $8 for every guest he invites. How many guests can there be
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Answer:

4 guests

Step-by-step explanation:

If it's $8 every person, 4 x $8 is $32, so he can have 4 guests.

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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.
Otrada [13]

I guess the "5" is supposed to represent the integral sign?

I=\displaystyle\int_1^4\ln t\,\mathrm dt

With n=10 subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

\ell_i=1+\dfrac{3(i-1)}{10}

and right endpoints are given by

r_i=1+\dfrac{3i}{10}

where 1\le i\le10.

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, \dfrac{4-1}{10}=\dfrac3{10}, and "bases" equal to the values of \ln t at both endpoints of each subinterval. The area of the trapezoid over the i-th subinterval is

\dfrac{\ln\ell_i+\ln r_i}2\dfrac3{10}=\dfrac3{20}\ln(ell_ir_i)

Then the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{20}\ln(\ell_ir_i)\approx\boxed{2.540}

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of \ln t at the average of the subinterval's endpoints, \dfrac{\ell_i+r_i}2. The area of the rectangle over the i-th subinterval is then

\ln\left(\dfrac{\ell_i+r_i}2\right)\dfrac3{10}

so the integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac3{10}\ln\left(\dfrac{\ell_i+r_i}2\right)\approx\boxed{2.548}

c. For Simpson's rule, we find a quadratic interpolation of \ln t over each subinterval given by

P(t_i)=\ln\ell_i\dfrac{(t-m_i)(t-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+\ln m_i\dfrac{(t-\ell_i)(t-r_i)}{(m_i-\ell_i)(m_i-r_i)}+\ln r_i\dfrac{(t-\ell_i)(t-m_i)}{(r_i-\ell_i)(r_i-m_i)}

where m_i is the midpoint of the i-th subinterval,

m_i=\dfrac{\ell_i+r_i}2

Then the integral I is equal to the sum of the integrals of each interpolation over the corresponding i-th subinterval.

I\approx\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt

It's easy to show that

\displaystyle\int_{\ell_i}^{r_i}P(t_i)\,\mathrm dt=\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)

so that the value of the overall integral is approximately

I\approx\displaystyle\sum_{i=1}^{10}\frac{r_i-\ell_i}6(\ln\ell_i+4\ln m_i+\ln r_i)\approx\boxed{2.545}

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Joan works on commission for a furniture store. She gets a base pay of $85 a day plus 6 percent of the
podryga [215]

Answer:

Correct expression: 85 + 0.06(t - 400)

Correct earnings: $91

Step-by-step explanation:

The expression Joan wrote is incorrect because instead, $85, she put $400 as her base salary. The correct expression would be 85 + 0.06(t - 400), where 85 is her base pay in dollars, and t - 400 represents her sales in dollars exceeding $400.

If Joan's sales were $500, she would not make $424.90, but

85 + 0.06(t - 400)

85 + 0.06(500 - 400)

85 + 0.06(100)

85 + 6

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Hope this helps.

Cheers,

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Answer:

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Step-by-step explanation:


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2x-y&lt;-3
timama [110]

Answer:

y > 2x + 3

Step-by-step explanation:

Okay, first let's simplify the equation by understanding the form.

y = mx + b is the form.

Currently we have 2x - y < -3, So we can subtract 2x on both sides to get

-y < -2x - 3

Let's multiply each term by -1 to get

y < 2x + 3, but wait! We have to flip the sign!

The answer is :

y > 2x + 3

5 0
3 years ago
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