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3 years ago

Plz helps brainlist and points also Q's in the photo its math

Mathematics

1 answer:

victus00 [196]3 years ago

8 0

answer :

A, C, and D

hope this helps:)

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There are 450 people at a play, the total receipt for $600 and the admission was $2 for adults and $0.75 for children, how many

sdas [7]

Answer:

340 children, 210 adults.

Step-by-step explanation:

$Let \ a \ - adults, \ c - children;\\c + a = 450\\0.75c + 2a = 600|\cdot 4\\3c + 8a = 2400;\\\\c + a = 450|\cdot(-3)\\3c + 8a = 2400\\-3c -3a = -1350\\5a = 1050 => a = 210, c = 450 -210 = 340.$

8 0

3 years ago

Here are scores for two softball teams for seven innings. Each team's total score is the sum of the numbers in its row. Rosie's

gladu [14]

Answer:

<h2> T<u>he Susans Bees lost by </u><u>3</u><u> points.</u></h2>

Step-by-step explanation:

Rosie's Riveters has 13 points in all.

The Susans Bees have 10 points in all.

13 - 10 = 3.

5 0

3 years ago

How much would $200 invested at 5% interest compounded monthly be worth after 9years? Round your answer to the nearest cent

Feliz [49]

Answer:

$311.20

Step-by-step explanation:

Here we are required to use the Compound interest formula for finding the Amount at the end of 9th year

The formula is given as

$A=P(1+\frac{r}{n})^{tn}$

Where ,

A is the final amount

P is the initial amount = $200

r is the rate of interest = 5% annual = 0.05

n is the frequency of compounding in a year ( Here it is compounding monthly) = 12

t is the time period = 9

Now we substitute all these values in the formula and solve for A

$A=200(1+\frac{0.05}{12})^{9\times 12}$

$A=200(1+0.00416)^{108}$

$A=200(1.00416)^{108}$

$A=200 \times 1.556$

$A=311.20$

Hence the amount after 9 years will be $311.20

4 0

3 years ago

An amusement park charges admission plus a fee for each ride. Admission plus 2 rides costs $10. Admission plus five rides costs

LenaWriter [7]

Answer:

The charge for admission is $6 and the charge for each ride is $2

Step-by-step explanation:

Let

x ----> the charge for admission

y ----> the charge for each ride

we have that

$x+2y=10$ -----> equation A

$x+5y=16$ -----> equation B

Solve the system by elimination

Subtract equation B from equation A

$x+2y=10\\-(x+5y)=-16\\---------\\2y-5y=10-16\\-3y=-6\\y=2$

Find the value of x

substitute the value of y in any equation

$x+2(2)=10$

$x+4=10$

$x=10-4$

$x=6$

therefore

The charge for admission is $6 and the charge for each ride is $2

8 0

3 years ago

How would I do this problem?

kenny6666 [7]

If it's a geometric sequence then:

$a_1=27;\ a_2=27\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{27}{36}=\dfrac{3}{4}=0.75\\\\a_{n+1}=a_nr\\\\a_3=27\cdot0.75=20.25\ CORRECT$

We calculate the fourth and fifth term of the sequence:

$a_4=a_3r\to a_4=20.25\cdot0.75=15.1875\\\\a_5=a_4r\to a_5=15.1875\cdot0.75=11.390625$

Answer:

In year 4 15.1875 animals.

In year 5 11.390625 animals.

7 0

3 years ago