All categories

3 years ago

What is the value of the expression 4(x - y) + 2 when x = 7 and y = 1?

Mathematics

2 answers:

Ksju [112]3 years ago

7 0

Answer:

Step-by-step explanation:

4(7-1) +2

4(6) + 2

24+ 2

lesya [120]3 years ago

3 0

Answer: 26

Step-by-step explanation:

4(x-y)+2

Lets plug in 7 for x and 1 for y

4(7-1)+2=4(6)+2

4*6=24

24+2=26

26 is the answer

You might be interested in

Pls help me solve this volume problem

Ostrovityanka [42]

Answer:

220

Step-by-step explanation:

First, find the area of the bottom prism.

5x10x3 = 150.

Then find the area of the top prism.

The trapezoid's area is 7, multiply this by 10.

Add your quantities together.

You get 220.

8 0

3 years ago

The answer should be a fraction

Vedmedyk [2.9K]

Ok so 40y2/50y3.... you are going to cancel out the common factor (10)

4y2/5y3..
now apply the exponent rule, which is : xa /xb = 1/ xb-a

so...y2/y3 = 1/ y3 - 2 = 1/y

ANSWER : 4/5y

5 0

3 years ago

Read 2 more answers

How to differentiate ?

Bas_tet [7]

Use the power, product, and chain rules:

$y = x^2 (3x-1)^3$

• product rule

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}$

• power rule for the first term, and power/chain rules for the second term:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}$

• power rule

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3$

Now simplify.

$\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}$

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

$\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) = \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)$

Differentiate both sides and you end up with the same derivative:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2$

7 0

3 years ago

[20 POINTS] Find the area of the rectangle

zvonat [6]

5 points

Hope this helps

8 0

3 years ago

Actorize Completely PQ+ PR - RQ -Q²

sp2606 [1]

Answer:

(Q + R)(P - Q)

Step-by-step explanation:

PQ + PR - RQ - Q² = (Q + R)(P - Q)

8 0

2 years ago