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2 years ago

Answer me pls I will mark brainliest worth 50 points. (sorry when I put 100 it goes to 50).

Mathematics

2 answers:

Anettt [7]2 years ago

6 0

Answer:

■ A random sample of dogs at different animal shelters in Abu Dhabi shows that 12 of the 60 dogs are puppies. Abu Dhabi's animal shelters collectively house 1,200 dogs each year. About how many dogs in all of the city’s animal shelters are puppies?

➢12/60 are puppies. 60 are the number of dogs, 12 is number of puppies.

If there are 1200 dogs, we focus on the 60. 1200/60= 20

Whatever we do to the denominator we do to the numerator, so since we can multiply 60 by 20 to get 1200; we need to multiply 12 by 20 to get the total number of dogs. Which is 240.

60/1200 = x/20 1200x /1200 =12/20 x =240

$TextFromYourMrEx$

#CarryOnLearning

mars1129 [50]2 years ago

4 0

Answer:

240

Step-by-step explanation:

1) First you set up a proportion.

2) You do 12/60=x/1,200.

3) You cross multiply and do 12x1,200. Your answer is 14400.

4) Then you do 14400/60=240

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If y=3x+1 were changed to y=x+1 how would the graph of the new function compare with the original?

Olegator [25]

F(x)=3x+1 (preimage)
g(x)=x+1 (image)
it is undergoing a reduction/compression with translation.
In general, a linear transformation is
g(x) = a*f(bx-h)+k
h=horizontal translation (right if h>0, left if h<0, note formula has minus sign)
k=vertical translation (up if k>0, down if k<0)
a=vertical stretching, (stretching if |a|>1, compression if |a|<1, also, if a<0, a reflection across the x-axis is performed)
b=horizontal stretching (|b|>0 compression, |b|<0 stretching, also, if b<0, a reflection across the y-axis is performed)

In this case,
g(x)=f(x/3), so it is a horizontal stretching.
Note that the y-intercept remains unchanged.

5 0

3 years ago

PLEASE HELP! i have by the end of the day to turn this in. PLEASE DONT GUESS

ivann1987 [24]

Answer: See step by step

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7 0

3 years ago

Read 2 more answers

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately <u>2020.022</u>

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$