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geniusboy [140]

2 years ago

13

What the answer fast now

2 answers:

podryga [215]2 years ago

7 0

Answer:

45°

Step-by-step explanation:

This is a special 6 - 6 - 6√2 right triangle with angle measures 45° - 45° - 90°

kumpel [21]2 years ago

5 0

Answer: m∠R = 45°

Step-by-step explanation:

$6^{2}\ +\ 6^{2} = 72$

$\frac{\left(6\right)}{\sqrt{72}}=0.7071067812$

$\sin^{-1}\left(\frac{\left(6\right)}{\sqrt{72}}\right)= 45$

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213 is 30% of what amount

WITCHER [35]

Start by calling the unknown amount 'x'. We can then set up a relationship to represent the question:

$213 = (30\%) \times x$
or
$213=0.3x$

and then solve for x, by dividing both sides of our equation by 0.3

$\frac{213}{0.3}=\frac{0.3x}{0.3}$
$710=x$

213 is 30% of 710

To check: $710 \times 0.3 = 213$

3 0

2 years ago

1. Ella has 7 rolls and 6 loose medium shirts. Each shirt costs $10. What is the value of her

weqwewe [10]

Answer:

the value of her medium shirts is $60 if Ella makes 100 more shirts her new value is $600

6 0

1 year ago

Simplify 7 to the 6th power times 7 to the 5th power

guapka [62]

The answer would be 7 to the 11th power

when multiplying exponents, just add the exponents together

6 + 5 = 11 :)

3 0

2 years ago

Find all roots: x^3 + 7x^2 + 12x = 0 <br> Show all work and check your answer.

Aliun [14]

The three roots of x^3 + 7x^2 + 12x = 0 is 0,-3 and -4

<u>Solution:</u>

We have been given a cubic polynomial.

$x^{3}+7 x^{2}+12 x=0$

We need to find the three roots of the given polynomial.

Since it is a cubic polynomial, we can start by taking ‘x’ common from the equation.

This gives us:

$x^{3}+7 x^{2}+12 x=0$

$x\left(x^{2}+7 x+12\right)=0$ ----- eqn 1

So, from the above eq1 we can find the first root of the polynomial, which will be:

x = 0

Now, we need to find the remaining two roots which are taken from the remaining part of the equation which is:

$x^{2}+7 x+12=0$

we have to use the quadratic equation to solve this polynomial. The quadratic formula is:

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Now, a = 1, b = 7 and c = 12

By substituting the values of a,b and c in the quadratic equation we get;

$\begin{array}{l}{x=\frac{-7 \pm \sqrt{7^{2}-4 \times 1 \times 12}}{2 \times 1}} \\\\{x=\frac{-7 \pm \sqrt{1}}{2}}\end{array}$

<em><u>Therefore, the two roots are:</u></em>

$\begin{array}{l}{x=\frac{-7+\sqrt{1}}{2}=\frac{-7+1}{2}=\frac{-6}{2}} \\\\ {x=-3}\end{array}$

And,

$\begin{array}{c}{x=\frac{-7-\sqrt{1}}{2}} \\\\ {x=-4}\end{array}$

Hence, the three roots of the given cubic polynomial is 0, -3 and -4

4 0

2 years ago

Thank you! Please do both

SVETLANKA909090 [29]

Answer:

my 12 inch pito that pito good

8 0

3 years ago