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vazorg [7]
3 years ago
10

What is the length of line PS

Mathematics
1 answer:
stiks02 [169]3 years ago
4 0
PQ^2 = PR \times PS

(4x + 4)^2 = (3x + 3)(3x + 3 + 21)

16x^2 + 32x + 16= (3x + 3)(3x + 24)

16x^2 + 32x + 16= 9x^2 + 81x + 72

7x^2 - 49x- 56 = 0

7(x^2 - 7x - 8)= 0

(x - 8)(x + 1) = 0

x - 8 = 0~~~or~~~x + 1 = 0

x = 8~~~or~~~x = -1

x = -1 makes the lengths of segments PQ and PR equal zero.
That is not possible, so the solution x = -1 is discarded.

x = 8

PS = 3x + 3 + 21

PS = 3x + 24

PS = 3(8) + 24

PS = 24 + 24

PS = 48
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a shopkeeper sold a certain number ( a two-digit number)of toys all priced at a certain value (also a two-digit number when expr
Makovka662 [10]

The answer is 91 toys sold, make the number ab where a is the 10th digit and b is the first digit. The value is 10a + b that can expressed as 10 (3) + 4 = 34

Let the price of each item: xy

10x + y

He accidentally reversed the digits to: 10b + a toys sold at 10y + x rupees per toy. To get use the formula, he sold 10a + b toys but thought he sold 10b + a toys. The number of toys that he thought he left over was 72 items more than the actual amount of toys left over. So he sold 72 more toys than he thought:

10a + b =10b + a +72

9a = 9b + 72

a = b + 8

The only numbers that could work are a = 9 and b = 1 since a and b each have to be 1 digit numbers. He reversed the digits and thought he sold 19 toys. So the actual number of toys sold was 10a + b = 10 (9) + 1 = 91 toys sold. By checking, he sold 91 – 19 = 72 toys more than the amount that he though the sold. As a result, the number of toys he thought he left over was 72 more than the actual amount left over as was stated in the question.

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3 0
3 years ago
9Find -18 + (-67). Show your work.
polet [3.4K]

Answer:

-85

Explanation:

Given the mathematical expression:

-18+(-67)

First, we recall the product of signs.

+\times-=-

So, first, we open the bracket:

-18+(-67)=-18-67

We then simplify:

=-85

The result is -85.

3 0
1 year ago
How do i evaluate (7 to the power of 2)to the power of 3)=7 to the power of 5​
kakasveta [241]

Step-by-step explanation:

(7^2)^3=7^5\qquad\bold{FALSE}\\\\\text{because}\\\\7^2=7\cdot7\\\\(7^2)^3=(7\cdot7)^3=(7\cdot7)\cdot(7\cdot7)\cdot(7\cdot7)=\underbrace{7\cdot7\cdot7\cdot7\cdot7\cdot7}_6=7^6\\\\\text{or use}\ (a^n)^m=a^{nm}\\\\(7^2)^3=7^{2\cdot3}=7^6

4 0
2 years ago
Read 2 more answers
Witch Expression gives the closest estimate of 228-179
torisob [31]

Answer: 230 - 180

Step-by-step explanation:

8 0
2 years ago
A certain forest covers an area of
iVinArrow [24]

Answer:

2598 square kilometers

Step-by-step explanation:

Hello

Step 1

year one

using a rule of three is possible to find how much is 8.75 od 4500 km2

Let

if

4500 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

4500:100\\x:8.75\\\frac{4500}{100}=\frac{x}{8.75}\\x=\frac{4500*8.75}{100} \\x=393.75\\

at the end of the year one, the area will be

4500-393.75=4106.25

this will be the initial area for the year 2.

Step 2

repite the step 1 with area initial =4106.25 km2

4106.25 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

4106.25:100\\x:8.75\\\frac{4106.25}{100}=\frac{x}{8.75}\\x=\frac{4106.25*8.75}{100} \\x=359.29\\

at the end of the year 2, the area will be

4106-359.29=3746.70

this will be the initial area for the year 3.

Step 3

repite the step 1 with area initial =4106.25 km2

3746.70 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3746.70:100\\x:8.75\\\frac{3746.70}{100}=\frac{x}{8.75}\\x=\frac{3746.7*8.75}{100} \\x=327.83\\

at the end of the year 3, the area will be

3746.70-327.83=3419.09

this will be the initial area for the year  4.

Step 4

year four

repite the step 1 with area initial =3419.09 km2

3419.09 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3419.09:100\\x:8.75\\\frac{3419.09}{100}=\frac{x}{8.75}\\x=\frac{3419.09*8.75}{100} \\x=299.17\\

at the end of the year 4, the area will be

3419.09-299.173=3119.82

this will be the initial area for the year  5.

Step 5

year five

repite the step 1 with area initial =3119.82 km2

3119.82 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

3119.82:100\\x:8.75\\\frac{3119.82}{100}=\frac{x}{8.75}\\x=\frac{3119.82*8.75}{100} \\x=272.99\\

at the end of the year 5, the area will be

3119.82-272.99=2846.92

this will be the initial area for the year  6.

Step 6

year six

repite the step 1 with area initial =2846.92km2

2846.92 km2  ⇒ 100$

x?km2         ⇒8.75

do the relation and isolate x

2846.92:100\\x:8.75\\\frac{2846.92}{100}=\frac{x}{8.75}\\x=\frac{2846.92*8.75}{100} \\x=249.10\\

at the end of the year six, the area will be

2846.92-249.10=2597.82 square kilometers

Have a great day.

8 0
3 years ago
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