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3 years ago

Help ASAP √3 (5 + √2)

Mathematics

2 answers:

FinnZ [79.3K]3 years ago

6 0

Answer:

5*sqrt(3)+sqrt(6)

Step-by-step explanation:

sqrt(3)*(5+sqrt(2))=5*sqrt(3)+sqrt(6)

Assoli18 [71]3 years ago

6 0

Answer:

(5 x sqrt(3)) + sqrt(6)

<em>Brainliest, please! (Almost an Ace!)</em>

Step-by-step explanation:

Distribute the sqrt(3) through and get:

(5 x sqrt(3)) + (sqrt(3) x sqrt(2))

Next, we simplify.

(5 x sqrt(3)) + sqrt(6)

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What is the slope of the linear function represented in the table?

DIA [1.3K]

Answer:

C. 1/7

Step-by-step explanation:

m = y2-y1/x2-x1

m = (1-0)/(0--7)

m = 1/7

4 0

3 years ago

WILL MARK BRAINLIEST!! <br><br> Solve for k.<br><br> -5 < -5k - 15 < 20

Tanzania [10]

Answer:

-2 > k > -7

Step-by-step explanation:

what you do to one section, you have to do to all

-5 < -5k - 15 < 20

add 15 to each part

10 < -5k < 35

divide by -5 (when you divide by a negative, it flips the inequality signs!)

-2 > k > -7

7 0

3 years ago

Determine the value of c that will result in a perfect square trinomial. <br> w^2+6w+ c =130+c

Genrish500 [490]

Answer:

The value of c that will result in a perfect square trinomial is (3)^2 or 9

The perfect square trinomial is: $(w+3)^2=139$

Step-by-step explanation:

We need to determine the value of c that will result in a perfect square trinomial.

$w^2+6w+ c =130+c$

Perfect square trinomial are of form: $a^2+2ab+b^2=(a+b)^2$

Now, the equation given is:

$w^2+6w+ c =130+c$

Looking at the term 6w, we can write it as 2(w)(3)

We are given: a = w, 2ab = 2(w)(3) so, b will be: (3)^2

So, we will be adding (3)^2 on both sides

$w^2+6w+ c=130+c\\w^2+2(w)(3)+ (3)^2 =130+(3)^2\\The\:left\:side\:becomes: a^2+2ab+b^2\\We\:can\:write\:it\:as: (a+b)^2\\We\:have\:a=w\: and\: b=3\\(w+3)^2=130+9\\(w+3)^2=139$

So, The value of c that will result in a perfect square trinomial is (3)^2 or 9

The perfect square trinomial is: $(w+3)^2=139$

3 0

3 years ago

Which is the equation of a hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0)? y squared over 40 minus x square

N76 [4]

The equation of the hyperbola with directrices at x = ±2 and foci at (5, 0) and (−5, 0) is $\frac{x^2}{10} + \frac{y^2}{15} = 1$

<h3>How to determine the equation of the hyperbola?</h3>

The given parameters are:

Directrices at x = ±2
Foci at (5, 0) and (−5, 0)

The foci of a hyperbola are represented as:

Foci = (k ± c, h)

The center is:

Center = (h,k)

And the directrix is:

Directrix, x = h ± a²/c

By comparison, we have:

k ± c = ±5

h = 0

h ± a²/c = ±2

Substitute h = 0 in h ± a²/c = ±2

0 ± a²/c = ±2

This gives

a²/c = 2

Multiply both sides by c

a² = 2c

k ± c = ±5 means that:

k ± c = 0 ± 5

By comparison, we have:

k = 0 and c = 5

Substitute c = 5 in a² = 2c

a² = 2 * 5

a² = 10

Next, we calculate b using:

b² = c² - a²

This gives

b² = 5² - 10

Evaluate

b² = 15

The hyperbola is represented as:

$\frac{(x - k)^2}{a^2} + \frac{(y - h)^2}{b^2} = 1$

So, we have:

$\frac{(x - 0)^2}{10} + \frac{(y - 0)^2}{15} = 1$

Evaluate

$\frac{x^2}{10} + \frac{y^2}{15} = 1$

Hence, the equation of the hyperbola is $\frac{x^2}{10} + \frac{y^2}{15} = 1$