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2 years ago

Estimate the product of 73 and 28

2 answers:

6 0

When I multiplied 73 and 28 I got 2,044 and estimated to 2,000. Then I multiplied 70 and 30 rounded those 2 numbers and got 2,100 thats the answer I got.

andreyandreev [35.5K]2 years ago

3 0

Round the numbers to 70 and 30, then multiply them together to get the product of 2100

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The temperatures were 62°F, 60°F, 59°F, 64°F, 65°F, 60°F, 70°F, 78°F, 60°F, and 71°F.

dmitriy555 [2]

Median- 63
mode-60
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6 0

2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20log_%7B2%7D%283x%20%2B%204%29%20%20-%207%20log_%7B4%7D%7Bx%7D%5E%7B2%7D%20%20%2B%20%20log_%

olga2289 [7]

First of all, we need all logarithms to have the same base. So, we use the formula

$\log_a(b)=\dfrac{\log_c(b)}{\log_c(a)}$

To change the second term as follows:

$\log_4(x^2)=\dfrac{\log_2(x^2)}{\log_2(4)}=\dfrac{\log_2(x^2)}{2}$

Finally, using the property

$\log(a^b)=b\log(a)$

we have

$\dfrac{\log_2(x^2)}{2}=\log_2(x)$

So, the equation becomes

$\log_2(3x+4)-7\log_2(x)+\log_2(x)=2 \iff \log_2(3x+4)-6\log_2(x)=2$

We can now use the formula

$\log(a)-\log(b)=\log\left(\dfrac{a}{b}\right)$

to write the equation as

$\log_2(3x+4)-6\log_2(x)=2 \iff \log_2(3x+4)-\log_2(x^6)=2 \iff \log_2\left(\dfrac{3x+4}{x^6}\right)=2$

Now consider both sides as exponents of 2:

$\dfrac{3x+4}{x^6}=4 \iff 4x^6-3x-4=0$

This equation has no "nice" solution, so I guess the problem is as simplifies as it can be

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3 years ago

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Monica [59]

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Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Help me please! No idea

zheka24 [161]

Answer:

Step-by-step explanation:

7 0

2 years ago

Find the parabola with equation y = ax^2 + bx whose tangent line at (2, 4) has equation y = 8x − 12.

MrRissso [65]

$\bf y=ax^2+bx\implies \cfrac{dy}{dx}=2ax+b$

now, we know the tangent line at (2,4) is y = 8x - 12, now, that's already in slope-intercept form, so, the slope of that tangent at (2,4) is "8" then.

now, that means when x = 2, the slope is 8, so let's use that.

$\bf \left. \cfrac{dy}{dx} \right|_{x=2}\implies 2a(2)+b=\stackrel{\textit{from tangent equation}}{8}\implies 4a+b=8
\\\\\\
4a=8-b\implies a=\cfrac{8-b}{4}\implies \boxed{a=2-\cfrac{b}{4}}\\\\
-------------------------------\\\\
\begin{cases}
x=2\\
y=4\\\\
a=2-\cfrac{b}{4}
\end{cases}\implies \stackrel{y = ax^2+bx}{4=\left( \boxed{2-\frac{b}{4}} \right)(2)^2+b(2)}\implies 4=8-b+2b
\\\\\\
\boxed{-4=b}\qquad thus\qquad a=2-\cfrac{-4}{4}\implies \boxed{a=3}\\\\
-------------------------------\\\\
y=3x^2-4x$

5 0

2 years ago