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3 years ago

1/15 times 7 in fraction form

Mathematics

2 answers:

Gwar [14]3 years ago

8 0

1/15 * 7

7/15 this is the answer in fraction form

jeka943 years ago

4 0

why hello there

Your answer is 7/15 you just multiple them and get 7/15

Sorry for the poor answer i know you expected more from me bur there is not much to explain.

If my answer did help please mark me as brainliest thank you and have a great day!

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A lorry driver charges $1800 to carry 60 tonnes of weight. How much will he charge for carrying 80 tonnes at the same rate

V125BC [204]

Answer:

$2400

Step-by-step explanation:

60 tonnes = $1800

80 tonnes = $?

= 1800 × 80/60

= $2400

4 0

3 years ago

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Please help me answer this question

lbvjy [14]

A. Let s be the smallest integer
B. We must add 2 to get the next greater integer
C. Second integer is s+2
D. s+(s+2)=118
2•s+2=118
2s+2-2=118-2
2s=116
2s/2=116/2
s=58
If the smallest even integer is 58 the consecutive integer is 58+2=60
Verify
58+60=118 ✔️

7 0

3 years ago

A women spent two thirds of her money. She lost two thirds of the remainder and then had $4 left. How much money did she start w

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3 years ago

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Expand (2x+2)^6 How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

1 year ago

RS = 4x – 9, ST = 19, RT = 80 – 14 Equation: 82 – 14 RS RT

algol13

Step-by-step explanation:

The expressions are not properly written

Given

RS = 4x – 9

ST = 19

RT = 8x – 14

Based on the given parameters, the addition postulate below is true

RS+ST = RT

Substitute

4x-9+19 = 8x-14

collect like terms

4x-8x = -14-19+9

-4x = -33+9

-4x = -24

x = -24/-4

x = 6

Get RS:

RS = 4x-9

RS = 4(6)- 9

RS = 24-9

RS = 15

Get RT:

RT = 8x - 14

RT = 8(6)-14

RT = 48-14

RT = 34

4 0

3 years ago