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2 years ago

Pleas help with the khan academy question

Mathematics

1 answer:

Alja [10]2 years ago

4 0

Answer:

1. 10

2. 4

3. 8

Step-by-step explanation:

You have to fill in the given values for y and undo in reverse order to find x.

1. y = 6

6 = x - 4

6 (+ 4) = x - 4 (+ 4)

10 = x

2. y = 0

0 = x - 4

0 (+ 4 ) = x - 4 (+ 4)

4 = x

3. y = 4

4 = x - 4

4 (+ 4) = x - 4 (+ 4)

8 = x

I hope this helps! Have a lovely day!! :)

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After 3.5 hours, Patrick had traveled 217 miles. If he travels at a constant speed, how far will he have traveled after 4 hours

Ede4ka [16]

Answer: He would have traveled 248 miles

Explanation: This is a ratio

To find how many miles he would have traveled after 4 hours, we need to apply the information givin into a ration table

(We can put miles on the top and hours on the bottom)

217 ?
------- ------
3.5 4

Now, we can do two things:
1. cross multiply
we would take 217 times 4 (868) and divide that by 3.5 (248)

2. Divide the top and bottom by 3.5 (numerator= 62 denominator= 1)
multiply 62 by 4 to get 248

Hope this helps!

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3 years ago

Find the missing coefficent. (5d - 7)(5d - 6) = 25d2 + d + 42

Tcecarenko [31]

25d^2-30d-35d+42=25d^2-65d+42.

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3 years ago

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A volleyball reaches its maximum of 13 feet, 3 seconds after it is served. What quadratic formula could model the height of the

Katena32 [7]

Answer:

$H=-(t-3)^2+13$

Step-by-step explanation:

The path covered by the volleyball will be a downward parabola with the vertex being the highest point of the ball.

A general form of a downward parabola is given as:

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

Where $(h,k)$ is the vertex of the parabola and 'a' is a constant.

Now, let 'h' be the vertical height and 't' be the time taken.

So, the equation would be of the form:

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Now, as per question:

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2 years ago

What is the 6th term of the geometric sequence where a1 = -4096 and a4 = 64?

Akimi4 [234]

$\bf \begin{array}{llccll}
term&value\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
a_1&-4096\\
a_2&-4096r\\
a_3&-4096rr\\
a_4&-4096rrr\\
&-4096r^3\\
&64
\end{array}\implies -4096r^3=64
\\\\\\
r^3=\cfrac{64}{-4096}\implies r^3=-\cfrac{1}{64}\implies r=\sqrt[3]{-\cfrac{1}{64}}
\\\\\\
r=\cfrac{\sqrt[3]{-1}}{\sqrt[3]{64}}\implies \boxed{r=\cfrac{-1}{4}}\\\\
-------------------------------$

$\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-\frac{1}{4}\\
a_1=-4096\\
n=6
\end{cases}
\\\\\\
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Artyom0805 [142]

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Step-by-step explanation:

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2 years ago

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