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4 years ago

Rose bought 7/20 of candy and 0.4 killogram of fruit which did she buy more of explain your answer

Mathematics

1 answer:

Juliette [100K]4 years ago

6 0

Simplify 7/20 to decimals by dividing.

7 divided by 20= 0.35

Rose bought 0.35 kilograms of candy and 0.4 kilograms of fruit.

So, she bought more fruit, since 7/20 is less than 0.4.

I hope this helps :)

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3 years ago

Are you supposed to use order of operations with a number sentence? (yes or no?)

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Yes with any math expression or equation always use order of operation

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4 years ago

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When completing the square on the equation c2 + 110 = 12, the resulting solution is: It

Delvig [45]

Answer:

$c=1$ and $c=-12$

Step-by-step explanation:

The correct equation is:

$c^2+11c=12$

To solve by completing the square method.

Solution:

We have:

$c^2+11c=12$

In order to solve by completing the square method we will carry out the following operations to the given equation to get a perfect square binomial.

We can write as:

$c^2+2.\frac{11}{2}c=12$

$c^2+2.\frac{11}{2}c+(\frac{11}{2})^2-(\frac{11}{2})^2=12$

$(c+\frac{11}{2})^2-(\frac{11}{2})^2=12$ [As $c^2+2.\frac{11}{2}c+(\frac{11}{2})^2=(c+\frac{11}{2})^2$ ]

Adding both sides by $(\frac{11}{2})^2$

$(c+\frac{11}{2})^2-(\frac{11}{2})^2+(\frac{11}{2})^2=12+(\frac{11}{2})^2$

$(c+\frac{11}{2})^2=12+\frac{121}{4}$

Taking LCD to add fraction.

$(c+\frac{11}{2})^2=\frac{48}{4}+\frac{121}{4}$

$(c+\frac{11}{2})^2=\frac{169}{4}$

Taking square root both sides.

$\sqrt{(c+\frac{11}{2})^2}=\sqrt{\frac{169}{4}}$

$c+\frac{11}{2}=\pm\frac{13}{2}$

Subtracting both sides by $\frac{11}{2}$ :

$c+\frac{11}{2}-\frac{11}{2}=\pm\frac{13}{2}-\frac{11}{2}$

$c=\pm\frac{13}{2}-\frac{11}{2}$

So, we have:

$c=\frac{13}{2}-\frac{11}{2}$ and $c=-\frac{13}{2}-\frac{11}{2}$

$c=\frac{2}{2}$ and $c=\frac{-24}{2}$

$c=1$ and $c=-12$ (Answer)

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4 years ago

Find the volume of this prism: ) 13cm 12cm 22cm 5cm

Arada [10]

Answer:

volume = 326.21

Step-by-step explanation:

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3 years ago

Over the closed interval [3,8] for which function can the extreme value theorem be applied?

vodka [1.7K]

The extreme value theorem can be applied on an interval if the function is continuous in the entire interval. Testing the continuity for each function, we get that the correct option is the third option.

Continuity:

A function f is continuous at an interval if all points in the interval are in the domain of the function, and, for each point of $x^{\ast}$ , the limit exists and:

$\lim_{x \rightarrow x^{\ast}} = f(x^{\ast})$

First function:

At $x = 4$ , the denominator is 0, and so, the extreme value theorem cannot be applied.

Second function:

At $x = 5$ , the denominator is 0, and so, the extreme value theorem cannot be applied.

Third function:

The only point there can be a discontinuity is at $x = 4$ , where the definition of the function changes. First we have to find the lateral limits, and if they are equal, the limits exist:

To the left(-), it is less than 4, so we take the definition for x < 4.

To the right(+), it is more than 4, so we take the definition for x >= 4.

$\lim_{x \rightarrow 4^{-}} h(x) = \lim_{x \rightarrow 4} \frac{9x}{10-x} = \frac{9*4}{10-4} = \frac{36}{6} = 6$

$\lim_{x \rightarrow 4^{+}} h(x) = \lim_{x \rightarrow 4} x + 2 = 4 + 2 = 6$

The lateral limits are equal, so the limit exists, and it's value is 6.

The definition at $x = 4$ is $h(x) = x + 2$ , so $h(4) = 4 + 2 = 6$ .

Since $\lim_{x \rightarrow 4} = h(4)$ , the function is continuous over the entire interval, and this is the correct answer.

Fourth function:

There can be a discontinuity at $x = 5$ , so we test the limits:

$\lim_{x \rightarrow 5^{-}} h(x) = \lim_{x \rightarrow 5} -x = -5$

$\lim_{x \rightarrow 5^{+}} h(x) = \lim_{x \rightarrow 5} x^2 - 20 = 5^2 - 20 = 25 - 20 = 5$

Different limits, so the limit does not exist and the function is not continuous at $x = 5$ , and the extreme value theorem cannot be applied.

For more on the extreme value theorem, you can check brainly.com/question/15585098

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3 years ago

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