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earnstyle [38]
3 years ago
15

Plz help plzzzzzzzz I’ll give u all mypoints

Mathematics
2 answers:
Svetach [21]3 years ago
6 0

<em>The picture sizes that are similar are </em><em>8*10 and 16*20. </em><em>Dilating the smaller picture size using a scale factor of </em><em>2</em><em> results in a larger picture size.</em>

DiKsa [7]3 years ago
5 0
A few of them are similar because of the sizes.
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Zoua’s older sister borrowed $3,000 from a bank to buy a used car. She asked Zoua to figure out the total amount she will have t
Fudgin [204]

Answer:

Answer is C

Step-by-step explanation:

hope you have s good day

3 0
3 years ago
Read 2 more answers
Identify the x-intercepts of the function below f(x)=x^2+12x+24
damaskus [11]

<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for (1) a = 1, b= 12 and c = 24

X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}

\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

8 0
3 years ago
WILL GIVE BRAINLIEST GIVE EXPLANATION
alex41 [277]

Answer:

5z+30

Step-by-step explanation:

5(z+6)

^ multiply each term in the parentheses by 5

5z+5x6

then,

5z+5x6 (multiply the numbers)

the answer will be 5z+30

5 0
3 years ago
In a shortage caused by a hurricane gas prices rose 15% it is now $2.80/gallon what was the price before
mash [69]

2.80*.15=.42

2.80-.42= 2.38

the answer is $2.38 (this was the price before)

8 0
3 years ago
(a) Use the definition to find an expression for the area under the curve y = x3 from 0 to 1 as a limit. lim n→∞ n i = 1 Correct
Luba_88 [7]

Splitting up the interval of integration into n subintervals gives the partition

\left[0,\dfrac1n\right],\left[\dfrac1n,\dfrac2n\right],\ldots,\left[\dfrac{n-1}n,1\right]

Each subinterval has length \dfrac{1-0}n=\dfrac1n. The right endpoints of each subinterval follow the sequence

r_i=\dfrac in

with i=1,2,3,\ldots,n. Then the left-endpoint Riemann sum that approximates the definite integral is

\displaystyle\sum_{i=1}^n\frac{{r_i}^3}n

and taking the limit as n\to\infty gives the area exactly. We have

\displaystyle\lim_{n\to\infty}\frac1n\sum_{i=1}^n\left(\frac in\right)^3=\lim_{n\to\infty}\frac{n^2(n+1)^2}{4n^3}=\boxed{\frac14}

6 0
3 years ago
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