Factor a number, variable, or expression out of the polynomial shown below. 15x^3 + 20x

Help me pls? :) ok thx

marishachu [46]

1 hour equals 70 miles so 70 times 4 equals 280 and half of 70 is 35 since you traveled 4.5 hours and 280+35 equals 315

8 0

3 years ago

A line passes through the point (10,5) and has a slope of 3/2. Write and equation in slope intercept form

ehidna [41]

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 10}}\quad ,&{{ 5}})\quad 
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2}
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-5=\cfrac{3}{2}(x-10)\implies y-5=\cfrac{3}{2}x-15
\\\\\\
y=\cfrac{3}{2}x-15+5\implies y=\cfrac{3}{2}x-10$

8 0

3 years ago

(10 POINTS) if you can answer ASAP

Ludmilka [50]

Answer:

A. -0.93

Step-by-step explanation:

It is a negative due to the scatter plot line decreasing as the precipitation increases, which cancels out C & D. Then, it is decreasing rapidly, which portrays it as closer to -1 than 0, meaning the answer is A.

8 0

3 years ago

Read 2 more answers

Determine what type of model best fits the given situation:

lyudmila [28]

Let value intially be = P

Then it is decreased by 20 %.

So 20% of P = $\frac{20}{100} \times P = 0.2P$

So after 1 year value is decreased by 0.2P

so value after 1 year will be = P - 0.2P (as its decreased so we will subtract 0.2P from original value P) = 0.8P-------------------------------------(1)

Similarly for 2nd year, this value 0.8P will again be decreased by 20 %

so 20% of 0.8P = $\frac{20}{100} \times 0.8P = (0.2)(0.8P)$

So after 2 years value is decreased by (0.2)(0.8P)

so value after 2 years will be = 0.8P - 0.2(0.8P)

taking 0.8P common out we get 0.8P(1-0.2)

= 0.8P(0.8)

$=P(0.8)^{2}$ -------------------------(2)

Similarly after 3 years, this value $P(0.8)^{2}$ will again be decreased by 20 %

so 20% of $P(0.8)^{2} \frac{20}{100} \times P(0.8)^{2} = (0.2)P(0.8)^{2}$

So after 3 years value is decreased by $(0.2)P(0.8)^{2}$

so value after 3 years will be = $P(0.8)^{2} - (0.2)P(0.8)^{2}$

taking $P(0.8)^{2}$ common out we get $P(0.8)^{2}(1-0.2)$

$P(0.8)^{2}(0.8)$

$P(0.8)^{3}$ -----------------------(3)

so from (1), (2), (3) we can see the following pattern

value after 1 year is P(0.8) or $P(0.8)^{1}$

value after 2 years is $P(0.8)^{2}$

value after 3 years is $P(0.8)^{3}$

so value after x years will be $P(0.8)^{x}$ ( whatever is the year, that is raised to power on 0.8)

So function is best described by exponential model

$y = P(0.8)^{x}$ where y is the value after x years

so thats the final answer

3 0

3 years ago

Can someone answer with steps and explanation? Thanks.

Vika [28.1K]

Answer:

A dilation by a factor of three about Point T followed by a translation of two units downwards.

Step-by-step explanation:

When transforming functions, we will reflect/dilate the figure first and then translate it. This is directly from the order of operations.

Since we are trying to determine the transformation that was performed, we can try to map ΔS'T'U' onto ΔSTU. We can start by translating the figure and then determining any reflections/dilations.

First, we can translate ΔS'T'U' up two units to map T' onto T. This is represented by the black triangle in the image below. Let the black triangle be ΔS''T''U''. (T'' and T are the same point.)

Next, notice that from Point T'' to U'', we move nine units right and six units up.

From Point T to Point U, we move three units right and two units up.

Likewise, from Point T'' to S'', we move six units left and nine units up.

From Point T to Point S, we move two units left and three units up.

Therefore, to map ΔS''T''U'' onto ΔSTU, we dilate ΔS''T''U'' about Point T by a factor of 1/3.

Hence, by reversing the transformations, to acquire ΔS'T'U', we can see that we will dilate ΔSTU by a factor of three about Point T and then a perform a translation of two units downwards.

8 0

3 years ago