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2 years ago

What is the area of this shape?

Mathematics

2 answers:

Akimi4 [234]2 years ago

4 0

Answer:

144

Step-by-step explanation:

The shape is a trapezoid.

Add 6 and 112, which is eighteen

divide eighteen by 2, which is 9

Finally, multiply 9 and sixteen which is 144

Eddi Din [679]2 years ago

4 0

Area =1/2(sum of parallel base) x (perpendicular height)

A=(1/2)(a+b)h
A=(1/2)(6+12)(16)
A= (1/2)(18)(16)
A=(9 x 16)
A=144 in2

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4 years ago

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Four families share a basket of 16 kilograms equally how many kilograms of apples does each family get?

alex41 [277]

Answer:

4 kg of apples

Step-by-step explanation:

Given that,

Number of families = 4

Weight of the basket of apples= 16 kg

We need to find how many kilograms of apples does each family gets. We can find it as follows :

Received = $\dfrac{16}{4}$ = 4 kg of apples

Hence, each family gets 4 kg of apples.

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3 years ago

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3 years ago

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

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3 years ago

Solve for x. Assume that lines which appear tangent are tangent. Help!!!

Serjik [45]

Answer:

x = 0

Step-by-step explanation:

From the intersecting secants theorem, If from an exterior point, we draw two secant segments to a circle, then the product of the length of one secant segment and its external secant part will be equal to the product of the length of the other secant segment and its external secant portion.

Thus, applying it to this question;

(x + 27 + 48) × (x + 27) = (x + 45)²

(x + 75)(x + 27) = x² + 90x + 2025

x² + 102x + 2025 = x² + 90x + 2025

Like terms will cancel put to give;

102x - 90x = 0

12x = 0

x = 0/12

x = 0

5 0

3 years ago