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3 years ago

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What is the area of the shaded region? DO NOT include the unit of measure in your answer

1 answer:

Aleks04 [339]3 years ago

3 0

The area is 73. You have to find the area of both triangles using the formula a = 1/2bh. The shaded triangle is 108 square yards but because another triangle is cut out of it, we need to subtract the area of the white triangle from the shaded triangle. The area of the white triangle is 35. So now we have 108-35 =73. That’s how i got 73.

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Quis<br>if<br>A= 3<br>B= 4<br>C= 2<br>so<br>A x B - C<br>is

Oksanka [162]

Answer:

if

A= 3

B= 4

C= 2

so

A x B - C

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3 x 4 - 2

=12 - 2

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7 0

3 years ago

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The capacity of a small bottle of glue is 185 ml. How much glue is there in 3 bottles?

Dmitrij [34]

You just need to multiply 185 by 3. 185 x 3 = 555

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3 years ago

What is the length of the major axis of the conic section shown below?

sveticcg [70]

$\dfrac{(x-2)^2}{49}+\dfrac{(y+5)^2}{36}=1$

The equation of a elipse:

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

The length of the major axis is equal 2a if a > b or 2b if b > a.

We have

$a^2=49\to a=\sqrt{49}=7\\\\b^2=36\to b=\sqrt{36}=6\\\\a > b$

therefore the length of the major axis is equal 2 · 7 = 14.

7 0

4 years ago

The 5th term in a geometric sequence is 160. The 7th term is 40. What are possible values of the 6th term of the sequence?

omeli [17]

Answer:

C. The 6th term is positive/negative 80

Step-by-step explanation:

Given

Geometric Progression

$T_5 = 160$

$T_7 = 40$

Required

$T_6$

To get the 6th term of the progression, first we need to solve for the first term and the common ratio of the progression;

To solve the common ratio;

Divide the 7th term by the 5th term; This gives

$\frac{T_7}{T_5} = \frac{40}{160}$

Divide the numerator and the denominator of the fraction by 40

$\frac{T_7}{T_5} = \frac{1}{4}$ ----- equation 1

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Where n is the nth term

So,

$T_7 = a r^{6}$

$T_5 = a r^{4}$

Substitute the above expression in equation 1

$\frac{T_7}{T_5} = \frac{1}{4}$ becomes

$\frac{ar^6}{ar^4} = \frac{1}{4}$

$r^2 = \frac{1}{4}$

Square root both sides

$r = \sqrt{\frac{1}{4}}$

r = ± $\frac{1}{2}$

Next, is to solve for the first term;

Using $T_5 = a r^{4}$

By substituting 160 for T5 and ± $\frac{1}{2}$ for r;

We get

$160 = a \frac{1}{2}^{4}$

$160 = a \frac{1}{16}$

Multiply through by 16

$16 * 160 = a \frac{1}{16} * 16$

$16 * 160 = a$

$2560 = a$

Now, we can easily solve for the 6th term

Recall that the formula of a GP is

$T_n = a r^{n-1}$

Here, n = 6;

$T_6 = a r^{6-1}$

$T_6 = a r^5$

$T_6 = 2560 r^5$

r = ± $\frac{1}{2}$

So,

$T_6 = 2560( \frac{1}{2}^5)$ or $T_6 = 2560( \frac{-1}{2}^5)$

$T_6 = 2560( \frac{1}{32})$ or $T_6 = 2560( \frac{-1}{32})$

$T_6 = 80$ or $T_6 = -80$

$T_6 =$ ±80

Hence, the 6th term is positive/negative 80

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3 years ago

Please help! Giveaway soon to celebrate expert level :D

elena-s [515]

Answer:

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3 years ago

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