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kap26 [50]
3 years ago
12

Find square root of 1.69

Mathematics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

The \sqrt{1.69} simplified is 1.3

Step-by-step explanation:

The square root of 1.69 is 1.3.

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there are 24 students in a class.8 of the students are boys.write the amount of boys in the class as a fraction in its simplest
Shkiper50 [21]

Answer:

8/24 which is 1/3

Hope that helps:) Brainliest?? Feel free to ask me more questions



Step-by-step explanation:


5 0
4 years ago
Read 2 more answers
HELP PLEASE QUICKLY!!! NO LINKS PLSS!!
dezoksy [38]
1a. (15x9)+(9x24)
135+216
351 ft squared
1b. 30 bags
3 0
3 years ago
Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so
IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

<em><u>Solution:</u></em>

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

<em><u>Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars</u></em>

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

14 \times c + 5 \times v=119

14c + 5v = 119 ------- eqn 1

<em><u>Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars</u></em>

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

10 \times c + 10 \times v = 130

10c + 10v = 130 -------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

<em><u>Subtract eqn 2 from eqn 3</u></em>

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

<em><u>Substitute c = 6 in eqn 1</u></em>

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0
3 years ago
What is the first step for solving the inequality 3y+14=44
uranmaximum [27]
3y+14=44

Answer: First step is to subtract 14 from each side

3y = 30 (divide each side by 3)
y = 30/3, y = 10
7 0
3 years ago
Need help now!! 20 points!!
Gelneren [198K]

When you have something like this, all you need to do is substitute the values, the last is for what value of x

For the first one;

((x^2+1)+(x-2))(2)

(x^2+x-1)(2)

(2)^2+(2)-1

4+2-1

5

For the second one;

((x^2+1)-(x-2))(3)

(x^2-x+3)(3)

(3)^2-(3)+3

9-3+3

9

For the last one;

3(x^2+1)(7)+2(x-2)(3)

3((7)^2+7)+2((3)-2)

3(49+7)+2(3-2)

3(56)+2(1)

168+2

170

3 0
3 years ago
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