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3 years ago

Four times the greatest of three consecutive odd

Mathematics

1 answer:

Rainbow [258]3 years ago

4 0

Answer:

The largest is 25.

Step-by-step explanation:

If the first integer is x, the second is x+1 and the third is x+2.

When you multiply the largest by 4: (x+2)*4

It is 31 more than 3 times the smallest: (x+2)*4 = (x)*3+31

When you simplify you get : 4x+8 = 3x+31

Which means: x = 23

But since you want the largest you get: 23+2 = 25

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mrs jones gave 1/5 of a cake to her neighbour. she then sliced the remainder into 3 pieces for her 3 children, ken, ron and davi

Arturiano [62]

The fraction of the cake that was ron's pieces is 1/5.

<h3>What is the fraction of the cake was ron's pieces?</h3>

Ratio demonstrates how many times one number can fit into another number. Ratios contrast two numbers by ordinarily dividing them. A/B will be the formula if one is comparing one data point (A) to another data point (B).

Since Mrs Jones gave 1/5 of a cake to her neighbour, the remaining fraction will be:

= 1 - 1/5

= 4/5

The fraction of the cake that was ron's pieces will be:

= 2/(1+3+4) × 4/5

= 2/8 × 4/5

= 8/40

= 1/5

Learn more about ratio on:

brainly.com/question/2328454

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4 0

1 year ago

When evaluating (6 x + 9) squared minus 5 for x = 1, the given value of 1 is substituted for what part of the expression?

STALIN [3.7K]

Answer:

the variable

Step-by-step explanation:

i got it right on edge :)

4 0

3 years ago

Read 2 more answers

If 100% of a number is 15, what is 50% of the number?

anzhelika [568]

Answer:

Let n be the number

100% of a number is 15 is written as: 100% × n = 15

100% = 100 / 100 = 1

Hence

100% × n = n = 15

We now need to find 50% of the number

50% of 15 = (50 / 100) × 15 = 750 / 100 = 7.5

Step-by-step explanation:

Mark me as brainliest pls

§ALEX§

6 0

4 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$