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xenn [34]
3 years ago
10

PARALLELOGRAM HELP ONLY ONE QUESTION TRUE OR FALSE

Mathematics
1 answer:
erik [133]3 years ago
7 0
The Circles Help Figure Out What Angle The Shape has
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* WILL GIVE BRAINIEST*
k0ka [10]

Answer:

B: 280

Step-by-step explanation:

The regression line predicts that when x equals 5:

log_{10}(y) = 2.447

In order to find the value for y, one must simply apply the following logarithmic property:

if : log_{b}(a) = c

then: b^c = a

Applying it to this particular problem:

log_{10}(y) = 2.2447\\10^{2.447}= y\\y=280

Therefore, the regression line predicts y will equal 280 when x equals 5.

3 0
3 years ago
Write an equation, in point-slope form, for the line that goes through points (4,4) and (-4,-1).
Soloha48 [4]

Answer:

C.) Y= 5/8x + 1 1/2

Step-by-step explanation:

To write the equation of a line, calculate the slope between points (4,4) and (-4,-1). After, substitute the slope and a point into the point slope form.

m = \frac{y_2-y_1}{x_2-x_1} = \frac{4--1}{4--4}= \frac{5}{8}

Substitute m = 5/8 and the point (4,4) into the point slope form.

y - y_1 = m(x-x_1)\\y -4 = \frac{5}{8}(x-4)\\y-4 = \frac{5}{8}x - \frac{20}{8} \\\y = \frac{5}{8}x - 2{1}{2} + 4 \\ y = \frac{5}{8}x + 1\frac{1}{2}

3 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
Josie is creating 5 music playlists for her dance classes. She will use a total of 60 songs. Each playlist will have 7 older son
melisa1 [442]

Answer: pls answer

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Algebra 2 help!!!!!!
Elza [17]
Yes here multiply takes place not divide
7 0
3 years ago
Read 2 more answers
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