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2 years ago

What is the value of p+9 when p=16

Mathematics

1 answer:

Tasya [4]2 years ago

4 0

Answer:

25 :)

Step-by-step explanation:

well, if it's p+9 and you know that p=16 then you plug 16 in for p so the equation will be 16+9 and then you solve from there which is 25

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How much cashews at $1.58 per pound must be mixed with walnuts that cost 78 cents per pound to make a mixture of 40 pounds of nu

Likurg_2 [28]

Number of pounds of cashews is 36 pounds and number of pounds of walnuts is 4 pounds.

Step-by-step explanation:

Step 1: Given details are cost of cashews = $1.58, cost of walnuts = 78 cents = $0.78, total weight of nuts = 40 pounds, cost of nuts per pound = $1.50

Step 2: Let number of pounds of cashews to be mixed be C, then number of pounds of walnuts will 40 - C. Form equation with these variables.

⇒ 1.58 C + (40-C) 0.78 = 40 × 1.50

⇒ 1.58 C + 31.2 - 0.78 C = 60

⇒ 1.58 C - 0.78 C = 60 - 31.2

⇒ 0.8 C = 28.8

⇒ C = 36 pounds

Step 3: Calculate number of pounds of walnuts

⇒ Number of pounds of walnuts = 40 - C = 4 pounds

5 0

2 years ago

Read 2 more answers

Solve the equastion for x + 2 1/2 =10

Virty [35]

Answer:

x= 7 1/2

Step-by-step explanation:

Subtract the 1/2 to get it out of the way (whatever you do on one side of the = sign you have to do to the other)

So 10- 1/2= 9 1/2

Subtract the 2 to get it out of the way

9 1/2-2=

7 1/2

Hope it helped :)

8 0

1 year ago

Help please who ever correct brainliest

Otrada [13]

Answer:

Step-by-step explanation:

3 _

7 0

3 years ago

For the funtion f(x)= (x+7)^5, Find f^-1 (x)

quester [9]

Answer:

The inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

Step-by-step explanation:

For the function $f(x)= (x+7)^5$ , Find $f^{-1} (x)$

For finding inverse of x,

First let:

$y=(x+7)^5$

Now replace x with y and y with x

$x=(y+7)^5$

Now, solve for y

Taking 5th square root on both sides

$\sqrt[5]{x}=\sqrt[5]{(y+7)^5}\\\sqrt[5]{x}=y+7\\=> y+7=\sqrt[5]{x}\\y=\sqrt[5]{x}-7$

Now, replace y with $f^{-1} (x)$

$f^{-1} (x)=\sqrt[5]{x}+7$

So, the inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

6 0

3 years ago

Pls solve question 4b. Verrry urgent . Ok tyy

kodGreya [7K]

Answer:

$y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}$

Step-by-step explanation:

Given equation:

$y = 3x^2 - 9x + 17$

Factor out 3 from the first 2 terms:

$y = 3(x^2 - 3x) + 17$

Divide the coefficient of x by 2 and square it: (-3 ÷ 2)² = 9/4

Add this inside the parentheses and subtract the distributed value of it outside the parentheses:

$y = 3\left(x^2 - 3x+\dfrac{9}{4}\right) + 17-\dfrac{27}{4}$

Factor the parentheses and combine the constants:

$y = 3\left(x-\dfrac{3}{2}\right)^2+\dfrac{41}{4}$

8 0

1 year ago

What is the value of p+9 when p=16​

What is the value of p+9 when p=16