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3 years ago

Find the distance between the origin and the point (-12,5). A. 12 B. 6 C. 7 D. 13

Mathematics

2 answers:

Neporo4naja [7]3 years ago

8 0

Answer:

I'd pick 13

Step-by-step explanation:

Let M (-12,5) be the given point and O (0,0) be the origin.

AlexFokin [52]3 years ago

7 0

Answer:

D. 13

Step-by-step explanation:

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(x-3y+ 3z=-4 2x+3y-z =15 (4x-3y- z=19 - What is the solution to the system?

lesya [120]

Answer:

x=5,y=1 and z=-2

Step-by-step explanation:

We are given that system of equation

$x-3y+3z=-4$ (I equation)

$2x+3y-z=15$ (II equation )

$4x-3y-z=19$ (III equation )

Equation II multiply by 3 then add with equation I

Then, we get

$7x+6y=41$ ....(Equation IV)

Subtract equation II from equation III then we get

$2x-6y=4$ (equation V)

Adding equation IV and equation V then, we get

$9x=45$

$x=5$

Substitute x=5 in equation V then, we get

$2(5)-6y=4$

$10-6y=4$

$6y=10-4=6$

$y=1$

Substitute x=5 and y=1 in equation then, we get

$5-3(1)+3z=-4$

$2+3z=-4$

$3z=-4-2=-6$

$z=\frac{-6}{3}=-2$

Hence, the solution for the given system of equation is given by

x=5,y=1 and z=-2

7 0

3 years ago

HELP ASAP!

ElenaW [278]

It is C because 5.69 multiplied by 19.69 equals 112.0361 Square Foot

6 0

4 years ago

Read 2 more answers

What would this problems answer be ?12y=2 1\4

mafiozo [28]

Answer:

Step-by-step explanation:12y ×4 =21

48y=21

Y=0.44

Mark brainlest if correct pls

5 0

4 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$