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babymother [125]
3 years ago
11

Zachary reads 6 books each month as part of his book club. if Zachary has read 24 books so far, how many months has he been with

his book club?​
Mathematics
2 answers:
DedPeter [7]3 years ago
8 0

Four months

Step-by-step explanation:

6+6=12

12+6=18

18+6=24

So, 6x4=24

vredina [299]3 years ago
5 0

Answer:

-6 books each month

-24 books

24 divided by 6

4 months

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A car is advertised as $288/month for 72 months. What is the cost of the car?
viktelen [127]

Answer:

$20,736

Step-by-step explanation:

When these kind of advertisements are displayed,, it means the EMI cost per month shall be $288.

Since it provides the time period, that is 72 months,

Total cost of car in this case shall be $288 \times 72 months = $20,736

Also, these include an interest factor, but overall it is = $20,736 only,

If one down payment for full price of the car is made the cost shall reduce by the interest amount, but since no interest rate is provided it shall be ignored.

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3 years ago
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Delvig [45]
The answer is 200\pi
3 0
3 years ago
Can someone pls explain and do this solution:(
Orlov [11]

Given,

SP~QP

SR~QR

In ∆SRP and ∆QRP

SP~QP (given)

SR~QR (given)

PR=PR (common)

Therefore,

∆SRP~∆QRP by SSS congruence rule.

6 0
2 years ago
Why do we use newton's<br>​
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2 years ago
Find the equation of the line tangent to the graph of
garik1379 [7]

Answer:

\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}

Step-by-step explanation:

We want to find the equation of the line tangent to the graph of:

\displaystyle y=\sin^{-1}\big(\frac{x}{5}\big)\text{ at } x=\frac{5}{2}

So, we will find the derivative of our equation first. Applying the chain rule, we acquire that:

\displaystyle y^\prime=\frac{1}{\sqrt{1-(\frac{x}{5})^2}}\cdot\frac{1}{5}

Simplify:

\displaystyle y^\prime=\frac{1}{5\sqrt{1-\frac{x^2}{25}}}

We can factor out the denominator within the square root:

\displaystyle y^\prime =\frac{1}{5\sqrt{\frac{1}{25}\big(25-x^2)}}

Simplify:

\displaystyle y^\prime=\frac{1}{\sqrt{25-x^2}}

So, we can find the slope of the tangent line at <em>x</em> = 5/2. By substitution:

\displaystyle y^\prime=\frac{1}{\sqrt{25-(5/2)^2}}

Evaluate:

\displaystyle y^\prime=\frac{1}{\sqrt{75/4}}=\frac{1}{\frac{5\sqrt{3}}{2}}=\frac{2\sqrt{3}}{15}

We will also need the point at <em>x</em> = 5/2. Using our original equation, we acquire that:

\displaystyle y=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}

So, a point is (5/2, π/6).

Finally, by using the point-slope form, we can write:

\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}(x-\frac{5}{2})

Distribute:

\displaystyle y-\frac{\pi}{6}=\frac{2\sqrt{3}}{15}x+\frac{-\sqrt{3}}{3}

Isolate. Hence, our equation is:

\displaystyle y=\frac{2\sqrt{3}}{15}x+\frac{\pi-2\sqrt{3}}{6}

7 0
2 years ago
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