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3 years ago

15

Which graph shows the equation v=4+2t

1 answer:

ehidna [41]3 years ago

5 0

You need to give options for future reference

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? cm<br> 3 cm<br> 7 cm<br> 4 cm<br> 7 cm<br> Missing length

Ket [755]

Answer:

5 cm

7cm

Step-by-step explanation:

8 0

3 years ago

A country is shaped like a trapezoid. It's northern border is about 9.6 miles across,and the southern border is approximately 25

Nostrana [21]

The area of the trapezoid is calculated through the equation,
A = 0.5(b₁ + b₂)h
where b₁ and b₂ are the bases and h is the height. Substituting the known values from the given,
A = 0.5(9.6 + 25)(90) = 1557 mi²
Therefore, the approximate area of the country is 1557 mi².

6 0

3 years ago

There are 520 marbles in a jar. The ratio of red to yellow marbles is 3:4 and the ratio of yellow to blue marbles is 2:3. What i

frutty [35]

Answer:

Step-by-step explanation:

Change the 2/3 to 4/6

Now the ratio becomes 3:4:6

So the number of each is

3x + 4x + 6x = 520

13x = 520

x = 40

Red = 3*40 = 120

Yellow = 4*40 = 160

Blue = 6 * 40 = 240

Total = 520

7 0

3 years ago

When dividing 878 by 31, a student finds a quotient of 28 with a remainder of 11. Check the students work, and use the check to

Liula [17]

878/3 is 28 with a remainder of 10! The error is when they subtracted 258-248 wrong

3 0

3 years ago

Read 2 more answers

Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

pantera1 [17]

This sequence has generating function

$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

5 0

3 years ago