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3 years ago

Find S15 for the series -1 + -3 + -5 + -7 +...

1 answer:

Ber [7]3 years ago

8 0

Hello :
the sum is arithmetic secquence of common -diff is :
(-7)-(-5) =(-5)-(-3) = (-3)-(-1) =-2 = d
the first term is : a1 = -1
the formula of the sum : S= (n/2)(a1 +an)
n is number of term an = a1+(n-1)d

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Write the equation in standard form for the circle that has a diameter with endpoints (6, -9)

nevsk [136]

Answer:

Step-by-step explanation:

By definition a diameter goes through the center of the circle. So if we find the midpoint of the diameter, we will have the h and k of the center. Then we will use one of the points ON the circle as x and y and solve for r. Here goes:

$M=(\frac{6+(-6)}{2},\frac{-9+(-9)}{2})$ to get a midpoint/center of (0, -9).

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

Filling in with h, k, x, and y:

$(6-0)^2+(-9-(-9))^2=r^2$ which simplifies to

$6^2+0^2=r^2$ so

$r^2=36$

Now we can write the equation of the circle:

$x^2+(y+9)^2=36$

7 0

3 years ago

Read 2 more answers

Helen's house is located on a rectangular lot that is1 1-8 miles by 9/10 mile.Estimate the distance around the lot

klemol [59]

Answer: The distance around the lot is given by

$4\frac{1}{20}\ miles$

Step-by-step explanation:

Since we have given that

Length of rectangular lot is given by

$1\frac{1}{8}\\\\=\frac{9}{8}\ miles$

Width of rectangular lot is given by

$\frac{9}{10}\ mile$

We need to find the distance around the lot.

As we know the formula for "Perimeter of rectangle":

$Perimeter=2(Length+Width)\\\\Perimeter=2(\frac{9}{8}+\frac{9}{10})\\\\Perimeter=2(\frac{45+36}{40})\\\\Perimeter=2\times \frac{81}{40}\\\\Perimeter=\frac{81}{20}\\\\Perimeter=4\frac{1}{20}\ miles$

Hence, the distance around the lot is given by

$4\frac{1}{20}\ miles$

3 0

3 years ago

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Given f(x), find g(x) and h(x) such that f(x)=g(h(x)) and neither g(x) nor h(x) is solely x. f(x)=−3x+4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√3+5

iVinArrow [24]

Answer:

$h(x)=-2x+4$

$g(x)=\sqrt[3]{x}-5$

Step-by-step explanation:

Consider the given function is

$f(x)=\sqrt[3]{-2x+4}-5$

It is given that $f(x)=g(h(x))$ and neither g(x) nor h(x) is solely x.

$f(x)=\sqrt[3]{(-2x+4)}-5$

Let $h(x)=-2x+4$ , then we get

$f(x)=g(h(x))=\sqrt[3]{h(x)}-5$

Substitute h(x)=x in the above function.

$g(x)=\sqrt[3]{x}-5$

Therefore, the required functions are $h(x)=-2x+4$ and $g(x)=\sqrt[3]{x}-5$ .

Check the solutions.

$g(h(x))=g(-2x+4)$ $[\because h(x)=-2x+4]$

$g(h(x))=\sqrt[3]{-2x+4}-5$ $[\because g(x)=\sqrt[3]{x}-5]$

$g(h(x))=f(x)$

Therefore, our solution is correct.

7 0

4 years ago

Teresa graphs the following 33 equations: y=2xy=2x, y=x2+2y=x2+2, and y=2x2y=2x2. She says that the graph of y=2xy=2x will event

Neko [114]

The correct question is
Teresa graphs the following 3 equations: y=2x, y=x2+2, and y=2x2. She says that the graph of y=2x will eventually surpass both of the other graphs. Is Teresa correct? Why or why not?

we have that
y=2x
y=x²+2
y=2x²

using a graph tool
see the attached figure

We can affirm the following
the three graphs present the same domain-----> the interval (-∞,∞)

The range of the graph y=2x is the interval (-∞,∞)

The range of the graphs y=x²+2 and y=2x² is the interval [0,∞)

therefore
Teresa is not correct because the graph of y = 2x will not surpass the other two graphs since in the interval of [0, infinite) the three graphs present the same range

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4 years ago

If y=4x+5 how much does the tree grow each year?

Elena L [17]

Answer:

4 ( not sure what the units are)

Step-by-step explanation:

Just by looking at the slope you can figure out how much it grows per year.

y = mx + b

m = slope

Not sure what units there are in the question just place it in the end.

Hope this helps :)

6 0

3 years ago

Read 2 more answers