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3 years ago

What is the slope of the line 3х + 5у — 2х + Зу

Mathematics

1 answer:

Shkiper50 [21]3 years ago

8 0

Answer:

-1/8

Step-by-step explanation:

You might be interested in

SA = 2tr2 + 2trh

julsineya [31]

Given:

Radius of the cylinder = 5 in

Height of the cylinder = 30 in.

To find:

The surface area of the cylinder.

Solution:

The surface area of the cylinder is:

$SA=2\pi r^2+2\pi rh$

Where, r is the radius and h is the height of the cylinder.

Putting $r=5,h=30,\pi=3.14$ in the above formula, we get

$SA=2(3.14)(5)^2+2(3.14)(5)(30)$

$SA=2(3.14)(25)+942$

$SA=157+942$

$SA=1099
$

Therefore, the surface area of the cylinder is 1099 square inches.

6 0

3 years ago

Solve for U 11=8+u/7.68

BlackZzzverrR [31]

$\begin{gathered}\\ \sf\longmapsto 11=\frac{8+u}{7.68}\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 11 = \frac{8 + u}{ \frac{192}{25} } \end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 11=\frac{(8+u)×25}{192}\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 11=\frac{200+25u}{192}\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 2112=200+25u\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto 2112-25u=200\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -25u=200-2112\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto -25u=-1912\end{gathered}$

$\begin{gathered}\\ \sf\longmapsto u=\frac{1912}{25}\end{gathered}$

Alternate form:

u=76.48

7 0

2 years ago

Show work and explain with formulas.

Juli2301 [7.4K]

20 Answer: a₁ = 4

Step-by-step explanation:

$a_n=324,\ r=3,\ n=5\\\\a_n=a_1 \cdot r^{n-1}\\\\324=a_1\cdot 3^{5-1}\\\\\dfrac{324}{3^4}=a_1\\\\\dfrac{324}{81}=a_1\\\\\large\boxed{4}=a_1$

21 Answer: n = 13

Step-by-step explanation:

$a_n=\dfrac{1}{64},\ a_1=64,\ r=\dfrac{1}{2}\\\\a_n=a_1 \cdot r^{n-1}\\\\\dfrac{1}{64}=64\cdot \bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{64\cdot 64}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{2^6\cdot 2^6}=\dfrac{1}{2^{n-1}}\\\\6+6=n-1\\\\\large\boxed{13}=n$

22 Answer: n = 5

Step-by-step explanation:

$a_n=48,\ a_1=1875\ r=\dfrac{2}{5}\\\\a_n=a_1 \cdot r^{n-1}\\\\48=1875\cdot \bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{48}{1875}=\bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{16}{625}=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\\bigg(\dfrac{2}{5}\bigg)^4=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\4=n-1\\\\\large\boxed{5}=n$