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3 years ago

What is 1 5/7 as a improper fraction?

Mathematics

2 answers:

Usimov [2.4K]3 years ago

7 0

12/7 is the answer to your problem

natulia [17]3 years ago

6 0

Improper Fraction
12/7 = 1 5/7

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How many of the first 15 positive integers can be written as the sum of three squares?

Jlenok [28]

Answer

no number from the following list can be expressed as the sum of three squares: 7, 15, 23, 31, 39, 47, ect... so only once

Step-by-step explanation:

nono dont touch me there this is my nono square

3 0

2 years ago

(04.01)

otez555 [7]

Answer:

A function can be any set of points where one input is equal to one output. You cannot have two X values with the same Y value, for example.

<u>I believe the answer is neither</u> because in both sets you can see they have different Y values for the X values 2 and 9.

6 0

3 years ago

michaela wrote her name in block letters. She colored 3/4 of the letters pink. She drew green dots on 1/2 of the pink letters. w

Vedmedyk [2.9K]

namely, what is 1/2 of 3/4, well, is just their product.

$\bf \stackrel{\textit{half of three quarters}}{\cfrac{3}{4}\cdot \cfrac{1}{2}}\implies \cfrac{3}{8}$

5 0

3 years ago

Plss help me and explaim too

masya89 [10]

Answer:

a = b = 1

Step-by-step explanation:

The triangles are similar thus ratios of corresponding sides are equal, that is

$\frac{10}{2}$ = $\frac{5}{b}$ ( cross- multiply )

10b = 10 ( divide both sides by 10 )

b = 1

and

$\frac{10}{2}$ = $\frac{5}{a}$ ( cross- multiply )

10a = 10 ( divide both sides by 10 )

a = 1

Thus a = 1 and b = 1

3 0

3 years ago

the half-life of strontium-90 is approximately 29 years. how much of a 500 g sample of strontium-90 will remain after 58 years

Yuliya22 [10]

Answer: 125 g

<u>Step-by-step explanation:</u>

$A = P_o\cdot e^{kt}\\\\\text{First, use the given information to find k:}\\\\\bullet A=\dfrac{1}{2}P_o\\\\\bullet k = unknown\\\\\bullet t=29\text{ years}\\\\\dfrac{1}{2}P_o=P_o\cdot e^{k(29)}\\\\\\\dfrac{1}{2}=e^{k(29)}\qquad divided\ both\ sides\ by\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=ln\bigg(e^{k(29)}\bigg)\qquad applied\ ln\ to\ both\ sides\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=29k\qquad simplified-ln\ and\ e\ cancel\ out\\\\\\\dfrac{ln\bigg(\dfrac{1}{2}\bigg)}{29}=k\qquad divided\ 29\ from\ both\ sides\\\\\\-0.0239=k$

$\text{Now, use the following in the equation to solve for A:}\\\\\bullet A=unknown\\\bullet P_o=500\\\bullet k=-0.0239\\\bullet t=58\text{ years}\\\\A=500\cdot e^{(-0.0239)(58)}\\\\.\quad=500\cdot e^{-1.386}\\\\.\quad=125$

8 0

3 years ago