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dimaraw [331]
3 years ago
7

V-10=-3 solve the equation

Mathematics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer:

v =7

Step-by-step explanation:

Simplifying

v + -10 = -3

Reorder the terms:

-10 + v = -3

Solving

-10 + v = -3

Solving for variable 'v'.

Move all terms containing v to the left, all other terms to the right.

Add '10' to each side of the equation.

-10 + 10 + v = -3 + 10

Combine like terms: -10 + 10 = 0

0 + v = -3 + 10

v = -3 + 10

Combine like terms: -3 + 10 = 7

v = 7

Simplifying

v = 7

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Is triangle RST is congruent to triangle XYZ? If so,name the postulate that applies.
inessss [21]
It definetly is congruent. 

Have a Great Day!!!
8 0
3 years ago
Read 2 more answers
In a completely randomized experimental design involving five treatments, 13 observations were recorded for each of the five tre
yawa3891 [41]

Answer:

MSE = 10

Step-by-step explanation:

Given

SSTR = 200

SST = 800

Required

Determine MSE

This is calculated as:

MSE = \frac{1}{ddf} * SSE

Where:

SSE = SST - SSTR

ddf \to denominator df

So, we have:

SSE = 800 - 200

SSE = 600

To calculate the df, we have:

r = 13 --- observations

n = 5 treatments

So:

ddf = Total\ df - Numerator\ df

Total = n*r-1 = 5*13 -1 = 64

Numerator =n - 1 = 5 - 1 =4

ddf =64-4=60

So, we have:

MSE = \frac{1}{ddf} * SSE

MSE = \frac{1}{60} * 600

MSE = 10

6 0
2 years ago
Can someone help me with this question? Thank you!
Harrizon [31]

\qquad\qquad\huge\underline{{\sf Answer}}

Let's evaluate ~

The given expression is :

\qquad \sf  \dashrightarrow \:4x + 8

plug the value of x as 3

\qquad \sf  \dashrightarrow \:4(3) + 8

\qquad \sf  \dashrightarrow \:12 + 8

\qquad \sf  \dashrightarrow \:20

Therefore, the required value is 20

8 0
2 years ago
Name a point that is between 50 and 60 units away from (7,-2) and state the distance between the two points.
laila [671]

Let the required point be (a,b)

The distance of (a,b) from (7,-2) is

= \sqrt{(a-7)^2+(b+2)^2}

But this distance needs to be betweem 50 & 60

So

50

Squaring all sides

2500 < (a-7)² + (b+2)² < 3600

Let a = 7

So we have

2500 < (b+2)² <3600

b+2 < 60    or  b+2 > -60   => b <58 or b > -62

Also

b+2 >50    or b + 2 < -50  => b >48 or B < -52

Let us take one value of b < 58 say b = 50

So now we have the point as (7, 50)

The other point is (7,-2)

Distance between them

= \sqrt{(7-7)^2+(50+2)^2}= \sqrt{(52)^2}=52

This is between 50 & 60

Hence one point which has a distance between 50 & 60 from the point (7,-2) is (7, 50)


6 0
3 years ago
Read 2 more answers
Consider the graph shown below. The maximum value of this function is.
AysviL [449]

Answer:

where is the graph please?

8 0
2 years ago
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