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Leona [35]
3 years ago
14

Quadrilateral RSTQ is a parallelogram.

Mathematics
1 answer:
Basile [38]3 years ago
3 0

<S ~= <Q AND TQ ~= SR

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6: REDUCE to simplest term. 4/1 + 15/2​
vovangra [49]

Answer:

\frac{4}{15}+\:\frac{1}{2} in the simplest term is:

\frac{4}{15}+\frac{1}{2}=\frac{23}{30}

Step-by-step explanation:

Given the expression

\frac{4}{15}+\:\frac{1}{2}

Thus, solving to reduce to the simplest term

\frac{4}{15}+\frac{1}{2}

LCM of 15, 2: 30

Adjusting fraction based on the LCM

so the expression becomes

\frac{4}{15}+\:\frac{1}{2}=\frac{8}{30}+\frac{15}{30}

Apply the fraction rule:      \frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}

            =\frac{8+15}{30}

Add the numbers: 8+15 = 23

            =\frac{23}{30}

Therefore, \frac{4}{15}+\:\frac{1}{2} in the simplest term is:

\frac{4}{15}+\frac{1}{2}=\frac{23}{30}

4 0
3 years ago
Slope of 0 3 and 4 1
attashe74 [19]

Answer:

m = \frac{-1}{2}

Step-by-step explanation:

The slope of (0,3) (4,1) =

\frac{-1}{2}

Hope this helps :)

6 0
3 years ago
Read 2 more answers
What is the approximate volume of a cone with a radius of 15 cm and a height of 4 cm? Round your answer to the nearest hundredth
vichka [17]

Answer:

Volume of a cone =942.86cm^3

Step-by-step explanation:

Given that the radius of a cone is 15cm and its height is 4cm

That is r=15cm and h=4cm

To find the volume of a cone :

volume of a cone=\frac{\pi r^2h}{3} cubic units

Now substitute the values in the formula we get

volume of a cone=\frac{(\frac{22}{7}) (15)^2(4)}{3}  

=\frac{(\frac{22}{7}) (225)(4)}{3}

=\frac{19800}{21}

=942.857

Now round to nearest hundredth

=942.86

Therefore Volume of a cone=942.86cm^3

6 0
3 years ago
What polynomial must be added to x^2-2x+6 so that the sum is 3x^2+7x?
slavikrds [6]
2x^2+9x-6 i found this by subtracting 3x^2+7x from the other equation
5 0
3 years ago
what is the correlation coefficient of the linear fit of the data shown below, to the nearest hundreth
8_murik_8 [283]

Answer:

theres no picture.

Step-by-step explanation:

5 0
3 years ago
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