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3 years ago

Please help for 12 points

Mathematics

1 answer:

ElenaW [278]3 years ago

4 0

Answer:

Step-by-step explanation:

The solution to a system of equations by graphing is found by graphing both equations on a coordinate plane. The solution is the point at which the lines intersect.

First, put both equations in slope-intercept form:

y = x + 4 Already in slope-intercept form

x - y = 1 Isolate y.

-y = -x + 1 Multiply both sides by -1.

y = x - 1

Next, graph both lines:

(See attachment)

In this situation, the lines don't intersect, meaning there is no solution to this system.

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Find the values of x and y

Zina [86]

Answer:

see below

Step-by-step explanation:

• find x

use tan rule

tan( 60 degrees) = x / 7

tan ( 60 ) x 7 = 12 . 1

x = 12. 1

• find y

you can either use Pythagoras theorem or the sin rule

let's use sin

sin ( 30 ) = 7 / y

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hope this helps, please mark it

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3 years ago

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Kipish [7]

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4 years ago

1st answer gets brainlyest Solve 3q - 5 = 13 for q.

almond37 [142]

Answer:

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3 years ago

Read 2 more answers

Determine the INTEREST ONLY for $4,331.00 earning 5.5% for 4 years, compounded semi-annually.

forsale [732]

4,331 times 8 (semiannual for 4 yrs) = 34,648
34,648 + 5.5% = 36,553.64

to find interest:
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THE INTEREST RATE IS $1,905.64

3 0

3 years ago

Find the equation of the plane passing through the points

erica [24]

Extract the normal vectors from the given planes:

$-2x+y+z+2=0\implies\vec n_1=(-2,1,1)$

$x+y-3z+1=0\implies\vec n_2=(1,1,-3)$

(which are unique up to their signs, meaning either $\vec n_1$ or $-\vec n_1$ are valid choices for the normal vector)

The third plane must be perpendicular to both these given planes, which means it would be parallel to both $\vec n_1$ and $\vec n_2$ , which in turn means its own normal vector $\vec n_3$ should be perpendicular to both $\vec n_1$ and $\vec n_2$ .