All categories

3 years ago

If 4x = 64, find the value of 2x + 3.

Mathematics

2 answers:

ipn [44]3 years ago

6 0

Answer:

Step-by-step explanation:

4x=64

divide that: =16

x=16

2(16)+3

bonufazy [111]3 years ago

6 0

Answer:

35 i think

Step-by-step explanation:

You might be interested in

4,3,2,5, 6, 6, 10, 5, 6, 2, 3, 4, 6, 7, 14,5 1. What is the mean of the data set?

tankabanditka [31]

Answer:5.5

Step-by-step explanation:

when all the numbers are added up =88

88 divide by 16 because their is 16 numbers in this sequence=5.5

3 0

3 years ago

Which of the following are factors of 11? A/1 B/11 C/4 D/5 E/0 F/22

7nadin3 [17]

Answer:

B and A

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Find the sum of the sequence. 33+34+35+36+...+130 I am unsure of how to solve this problem.

nadezda [96]

There are 98 terms.
98*(33+130)/2 = 7987

3 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

Determine the slope of the linear function. Y= 1/5 x + 6

Novay_Z [31]

D. 1/5

Reason: it’s always the first number in the equation

7 0

3 years ago

Read 2 more answers