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zavuch27 [327]
3 years ago
5

Natasha, Mark and Henry share some sweets in the ratio 5:4:4. Natasha gets 35 sweets. How many more sweets does Natasha get over

Henry?
Mathematics
1 answer:
Trava [24]3 years ago
3 0

Answer:

7 sweets

Step-by-step explanation:

This is because the constant of proportionality is 7 since 35/5 since it is the ratio for Natasha. Since 7 is the constant of proportionality, you multiply it by the difference of the ratios. In this case, Natasha and Henry have a difference of 1 in ratios. Since it is a difference of 1, you multiply that by the constant of proportionality, 7. 7 X 1 is equal to 7, so Natasha gets 7 more sweets than Henry.

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Describe the end behavior of the following function: F(x)=2x^4+x^3
Anika [276]

Answer:

Rises to the left and rises to the right.

Step-by-step explanation:

Since, the given function is f(x)=2x^{4}+x^{3}, and the end behavior of the given function is determined as:

Consider the given function f(x)=2x^{4}+x^{3}, identify the degree of the function:

The degree of the function is : 4 which is even

And then identify the leading coefficient of the given function that is +2 which is positive in nature.

Hence, the function is positive and even in nature, therefore, the end behavior of the function will be rising to the left and rising to the right.

3 0
3 years ago
What is the radical form of the expression (2x^4y^5)^3/8
s344n2d4d5 [400]

We are given expression: (2x^4y^5)^{3/8}

Let us distribute 3/8 over exponents in parenthesis, we get

(2^{3/8}x^{4\times 3/8}y^{5\times 3/8}) = (2^{3/8}x^{12/8}y^{15/8})

= (2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} )

We can get x and y out of the radical because, we get whlole number 1 for x and y exponents for the mixed fractions.

So, we could write it as.

(2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} ) = xy(2^{\frac{3}{8} }x^{\frac{4}{8}} y^{\frac{7}{8}} )

Now, we could write inside expression of parenthesis in radical form.

xy\sqrt[8]{2x^{3}x^4y^7}

8 0
3 years ago
3 1 - 2.6<br>________<br>0.05<br>​
olya-2409 [2.1K]

Work:

\frac{31 - 2.6}{0.05}

\frac{28.4}{0.05}

Answer:

567. \frac{}{9}

6 0
3 years ago
Of the 10,000 participants in the Cooper Run/Walk, 42% were runners and one-third of the runners were female. Sally, a female ru
Aleks [24]

Answer: 140 female runners

Step-by-step explanation:

First find out the number of runners:

= 42% * 10,000

= 4,200 runners

Then find out the number of female runners:

= 4,200 * 1/3

= 1,400 female runners

Sally finished faster than 90% of the female runners which means that 10% finished faster or at the same time as her.

= 10% * 1,400

= 140 female runners

5 0
3 years ago
Looking at the two quadratic functions below (1 &amp; 2), answer the following questions.
algol [13]
Part A:

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that parabola (2) is stretched horizontally by a factor of 13 which is greater than 1. This means that parabola (2) is further away from the x-axis than parabola (1). (i.e. parabola (2) is more 'vertical' than parabola (1).

Therefore, parabola (1) is wider than parabola (2).



Part B:

A parabola open up when the coefficient of the quadratic term (the squared term) is positive and opens down when the coefficient of the quadratic term is negative.

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that the coefficient of the quadratic term is positive for parabola (2) and negative for parabola (1), therefore, the parabola that will open down is parabola (1).



Part C:

For any function, f(x), the graph of the function is moved p places to the left when p is added to x (i.e. f(x + p)) and moves p places to the right when p is subtracted from x (i.e. f(x - p)).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (1), 12 is added to x, which means that the graph of the parent function is shifted 12 places to the left while in parabola (2), 4 is subtracted from x, which means that the graph of the parent function is shifted 4 places to the right.

Therefore, the parabola that would be furthest left on the x-axis is parabola (1).



Part D:

For any function, f(x), the graph of the function is moved q places up when q is added to the function (i.e. f(x) + q) and moves q places down when q is subtracted from the function (i.e. f(x) - q).

Given parabola (1) to be f(x) =-(x+12)^2 -6, and parabola (2) to be f(x)=13(x-4)^2+1

Notice that in parabola (2), 1 is added to the function, which means that the graph of the parent function is shifted 1 place up while in parabola (1), 6 is subtracted from the function, which means that the graph of the parent function is shifted 6 places down.

Therefore, the parabola that would be highest on the y-axis is parabola (2).
7 0
3 years ago
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