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3 years ago

13

Are there any maximums or minimums on the function

le="\sqrt[3]{x}" alt="\sqrt[3]{x}" align="absmiddle" class="latex-formula"> ?

1 answer:

iren2701 [21]3 years ago

6 0

No, there are no maximums or minimums for this function as all negative and positive numbers have cube roots.

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X2-3xy-5xy-7y2+4x2+8y2

Zepler [3.9K]

For this case we must simplify the following expression:

$x ^ 2-3xy-5xy-7y ^ 2 + 4x ^ 2 + 8y ^ 2$

We must add similar terms taking into account that:

Equal signs are added and the same sign is placed.

Different signs are subtracted and the major sign is placed.

$(x ^ 2 + 4x ^ 2) + (- 3xy-5xy) + (- 7y ^ 2 + 8y ^ 2) =\\5x ^ 2-8xy + y ^ 2$

Thus, the simplified expression is:

$5x ^ 2-8xy + y ^ 2$

Answer:

$5x ^ 2-8xy + y ^ 2$

4 0

3 years ago

What is the solution of the equation 70 = 1.4y ?

sdas [7]

So you basically have to divide 1.4 on both sides so you can get y by its self. Once you divide 60 by 1.4 you’ll get 50.

y=50

7 0

2 years ago

Use the Rules of Special right triangles to find x and y 6) у 30° 4 [ Select] [ Select] y =

storchak [24]

Answer:6.6 y

Step-by-step explanation:

6 0

2 years ago

A city council consists of seven Democrats and five Republicans. If a committee of four people is selected, find the probability

Hoochie [10]

Answer:

The probability of selecting two Democrats and two Republicans is 0.4242.

Step-by-step explanation:

The information provided is as follows:

A city council consists of seven Democrats and five Republicans.
A committee of four people is selected.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!\times (n-k)!}$

Compute the number of ways to select four people as follows:

${12\choose 4}=\frac{12!}{4!\times (12-4)!}=495$

Compute the number of ways to selected two Democrats as follows:

${7\choose 2}=\frac{7!}{2!\times (7-2)!}=21$

Compute the number of ways to selected two Republicans as follows:

${5\choose 2}=\frac{5!}{2!\times (5-2)!}=10$

Then the probability of selecting two Democrats and two Republicans as follows:

$P(\text{2 Democrats and 2 Republicans})=\frac{21\times 10}{495}=0.4242$

Thus, the probability of selecting two Democrats and two Republicans is 0.4242.

6 0

2 years ago

Help!!! I don’t get this

Goshia [24]

i think u gotta multiply

3 0

2 years ago

Read 2 more answers