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2 years ago

(06.04 MC)

Mathematics

1 answer:

ratelena [41]2 years ago

7 0

Answer:

54 sq units

Step-by-step explanation:

The triangle at the top (FAB) has a base of 10 and a height of 3. Area of triangle is 1/2bh so 1/2(10 * 3) = 1/2 * 30 = 15 sq units

If you cut off the triangle you have a trapezoid, EBCD. A trapezoid is (b1 + b2)/2 * h, so (6 + 7)/2 * 6 = 13/2 * 6 = 6.5 * 6 = 39 sq units

Add them together: 15 + 39 = 54 sq units

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X + 6 = −x − 6 (so i can ask)kjhgfdccfghjk

ratelena [41]

Answer:

x = -6

Step-by-step explanation:

Equations

Solve the equation:

x + 6 = -x - 6

We must find the value of x that makes the identity above true.

Let's join all the variables on the left side and the numbers on the right side.

Adding x:

x + 6 + x = -x - 6 + x

The variables cancel out on the right side:

2x + 6 = -6

Subtracting 6:

2x + 6 - 6 = -6 -6

The 6 and -6 are canceled out:

2x = -12

Dividing by 2:

x = -12/2

x = -6

8 0

2 years ago

Name a 2 digit odd number that is prime.name a 2 digit odd number that is compostite

e-lub [12.9K]

Prime for 2 digit odd number: 13 , 11 , 17, etc

Odd number that is composite: 39, 25, 15, etc

Hope this helps

7 0

2 years ago

PLEASE HELP

algol13

Independent events mean that one event happening does not have to have the other event to happen.

The probability of both events happening is the product of the events.

The answer would be:

E. P(A ∩ B) = P(A) x P(B)

8 0

2 years ago

Read 2 more answers

What is the second term of (s+v)^5 a)25s^4 b)25s^4v c)5s^4v d)125s^4v

mrs_skeptik [129]

By pascal triangle this second term should be
5s^4v

6 0

2 years ago

Read 2 more answers

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

7 0

2 years ago