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3 years ago

Someone help me with c ._.

Mathematics

1 answer:

Elis [28]3 years ago

4 0

Answer:

244 inches squared

Step-by-step explanation:

basically just subtract the area of four rectangles by the total area of cardboard. so you would get 500-256=244

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Can someone please help me I really need help please help me thank you

Scilla [17]

A shape with 4 angles always have 360 degrees altogether.

Which means:
43 + 167 + 90 + y = 360
y = 360 - 43 - 167 - 90
y = 60

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2 years ago

Evaluate x³ for x=2.

loris [4]

Answer:

Step-by-step explanation:

If we have anything to the third power, we are multiplying the number by itself 3 times.

If x = 2, then the expression is $2^3$ .

$2\cdot2\cdot2=8$

Hope this helped!

7 0

3 years ago

Read 2 more answers

A model airplane has a scale of 2 in: 9 yards. If the wing span of the model airplane is 18 inches, how many yards is the actual

mote1985 [20]

Answer:

81 yards

Step-by-step explanation:

8 0

3 years ago

If 5% more than a is b, and b is 15% less than, what is the ratio of a to c? Please answer quickly.

Cerrena [4.2K]

Answer: 24 5 + 9 = 15 + 9 = 24

Step-by-step explanation:

7 0

3 years ago

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Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive

Nostrana [21]

Answer:

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$ = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

$\sum_{k=0}^\infty a(r)^k$

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is $\sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}$

Given series,

$\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

$=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......$

The first four terms of the series are

$(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})$

Let

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$ and $t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

Now for $S_n$ ,

$S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......$

$=\sum_{n=0}^\infty9(\frac 17)^n$

It is a geometric series.

The common ratio of $S_n$ is $\frac17$

The sum of the series

$S_n=\sum_{n=0}^\infty \frac{9}{7^n}$

$=\frac{9}{1-\frac17}$

$=\frac{9}{\frac67}$

$=\frac{9\times 7}{6}$

=10.5

Now for $t_n$

$t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......$

$=\sum_{n=0}^\infty3(\frac 15)^n$

It is a geometric series.

The common ratio of $t_n$ is $\frac15$

The sum of the series

$t_n=\sum_{n=0}^\infty \frac{3}{5^n}$

$=\frac{3}{1-\frac15}$

$=\frac{3}{\frac45}$

$=\frac{3\times 5}{4}$

=3.75

The sum of the series is $\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}$

= $S_n+t_n$

=10.5+3.75

=14.25

4 0

3 years ago

Someone help me with c ._.​

Someone help me with c ._.