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borishaifa [10]
3 years ago
12

Can u help me pls they are exponents with multiplication and division

Mathematics
1 answer:
Ronch [10]3 years ago
3 0
When there is a multiplication involving exponents, we need to evaluate the numbers and each variables separately. In this case, we have 5 and 4 as the numbers, and w and h for the variables.

When we multiply 5 and 4 we get:
5 \times 4 = 20

For w, we have {w}^{6}, {w}^{4}, w. While we are multiplying the w's, we need to add the exponents together to get the final answer: {w}^{6} \times {w}^{4} \times w = {w}^{11}

For h's, there are {h}^{2}, {h}^{5}, {h}^{4}. When we apply the same thing to h's, we get {h}^{2} \times {h}^{5} \times {h}^{4} = {h}^{11}

For the final step, we need to write all of these together: 20 \times {w}^{11} \times {h}^{11} \: \: or \: \: 20 {w}^{11} {h}^{11} \: \: or \: \: 20 {(wh)}^{11}
You might be interested in
Lexi picked 2 7/10 pounds of strawberries. That afternoon, Lexi’s sister ate 7/8 of the strawberries. How many pounds of strawbe
gavmur [86]

3\frac{3}{35} pound of strawberry

Step-by-step explanation:

Given parameters:

    Pounds of strawberry picked by Lexi =2 \frac{7}{10}

    Pounds of strawberry eaten by Lexi sister = \frac{7}{8}

Unknown:

Pound of strawberry Lexi's sister ate = ?

Solution:

This is a fraction word problem.

  Let pound of strawberry eaten by Lexi's sister = K

We can establish that:

  Pound of strawberry eaten by Lexi sister x K = Pound of strawberry picked by Lexi

Mathematically we have:

         \frac{7}{8} x K = 2 \frac{7}{10}

Learn more:

fractions brainly.com/question/1757979

#learnwithBrainly

             K =  \frac{27 x 8}{10 x 7} = \frac{216}{70}

            K = 3\frac{3}{35} pound of strawberry

8 0
3 years ago
The revenue R you receive for selling pizza slices depends on the price p that you charge per slice and is modeled by R =-16p
sp2606 [1]

Answer:

p \geq 0

Step-by-step explanation:

Given

R(p) = -16p^2 + 80p + 5

Required

Determine the domain of the function

To do this, we need to solve for the vertex, p of the function

p= \frac{-b}{2a}

Given the the general form of a quadratic function is:

y = ax^2 + bx + c

By comparison, we have:

a = -16    b =80    c = 5

So:

p= \frac{-b}{2a}

p = \frac{-80}{-16 * 5}

p = \frac{-80}{-80}

p = 1

Substitute 1 for p in R(p) = -16p^2 + 80p + 5

R(1) = -16(1)^2 + 80(1) + 5

R(1) = -16 + 80 + 5

R(1) = 69

This implies that

(p, R) = (1,69)

The interpretation of this is that;

<em>For every value of p, there's a corresponding value of R.</em>

<em>However, because p indicates price and it's impossible to have a negative price, we can say that the minimum value of p is 0;</em>

Hence, the domain is

p \geq 0

8 0
3 years ago
Passing through (-1, -7) and perpendicular to the line with equation 4x + 5y = 31
MArishka [77]
4x + 5y = 31
5y = -4x + 31
y = -4/5x + 31/5...the slope here is -4/5. A perpendicular line will have a negative reciprocal slope. All that means is " flip " the slope and change the sign. So the slope we need is 5/4 (see how I flipped the slope and changed the sign)

y = mx + b
slope(m) = 5/4
(-1,-7)...x = -1 and y = -7
now we sub and find b, the y int
-7 = 5/4(-1) + b
-7 = -5/4 + b
-7 + 5/4 = b
-28/4 + 5/4 = b
- 23/4 = b

so ur perpendicular line will be : y = 5/4x - 23/4
4 0
3 years ago
I just need a little help​
iris [78.8K]

Answer:  

your answer should be 2 and 1/4ths

Step-by-step explanation

2 and 1/4ths bc 5 is more than 4 and there is already a whole so there for there are going to be 2 wholes then there are still parts left over and its out of 4 and there is only one left over so there for it would be 1/4ths.

7 0
3 years ago
The bearing of a plane from the airport is 65.
Ulleksa [173]

Answer:

The bearing of the airport from the plane is 245°

Step-by-step explanation:

The bearing of a point from a location is given by the angle in degrees of rotation from the Northern direction of the location to the direction of the location of the point

The given bearing of the plane from the airport = 65°

Therefore, the direction of the plane from the Northern direction of the airport = 65°

From the Northern direction of the plane, the bearing of the airport = 245°

5 0
3 years ago
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