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3 years ago

Quadrilateral with no parallel sides

Mathematics

1 answer:

taurus [48]3 years ago

4 0

Answer:

The answer is a trapezoid.

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A computer store has certain software on sale at 3 for $20.00, with a limit of 3 at the sale price. Additional

alexgriva [62]

Answer:

a) y = 9.95(x-3) + 20.00

b) Cost of 5 software packages = $39.90

Step-by-step explanation:

Given that:

Cost of 3 software at sale = $20

Cost of regular = $9.95

Let,

x be the number of software purchased.

We will subtract 3 from x because they cost $20.

y be the total cost.

y = 9.95(x-3)+ 20.00

Putting x=5 in expression

y = 9.95(5-3) + 20.00

y = 2(9.95) + 20.00

y = 19.90 + 20.00

y = $39.90

Hence,

a) y = 9.95(x-3) + 20.00

b) Cost of 5 software packages = $39.90

7 0

3 years ago

How do you solve -10+2y=7

Eva8 [605]

Answer

y = $\frac{17}{2}$

Step-by-step explanation:

-10 + 2y = 7

Add 10 to both sides of the equation

2y = 7 + 10

Add 7 and 10

2y = 17

divide each term in 2y=17 by 2

$\frac{2y }{2} = \frac{17}{2}$

Divide y by 7

y = $\frac{17}{2}$

Hope this helps

5 0

3 years ago

Find the value of x. Round to the nearest degree.

lord [1]

hii! how are you? hope you're having a good day! (: i wish you the best. <3 stay strong and stay safe!

5 0

3 years ago

Without a calculator how would you solve this?

mina [271]

Answer:

the largest number that is less than (2+√3)^6 is 2701

Step-by-step explanation:

$\bigstar \ (2+\sqrt{3})^6= \\\\C{}^{0}_{6}2^{6}+C{}^{1}_{6}2^{5}\sqrt{3}^1+C{}^{2}_{6}2^{4}\sqrt{3}^2+C{}^{3}_{6}2^{3}\sqrt{3}^3+C{}^{4}_{6}2^{2}\sqrt{3}^4+C{}^{5}_{6}2^{1}\sqrt{3}^5+C{}^{6}_{6}\sqrt{3}^6$

$=64+192\sqrt{3}+720 +480\sqrt{3} +540+108\sqrt{3} +27\\=1351+780\sqrt{3}$

$\bigstar\ (2-\sqrt{3} )^6 =\\\\1351-780\sqrt{3}$

$\bigstar\ \bigstar\ (2+\sqrt{3} )^6+(2-\sqrt{3} )^6 =\\\\1351+780\sqrt{3}+1351-780\sqrt{3}$

$\Longrightarrow(2+\sqrt{3} )^6+(2-\sqrt{3} )^6 =2702$

$\Longrightarrow(2+\sqrt{3} )^6 =2702-(2-\sqrt{3} )^6$

$0 < \left( 2-\sqrt{3} \right) < 1 \Longrightarrow 0 < \left( 2-\sqrt{3} \right)^{6} < 1 \Longrightarrow-1 < \left( 2-\sqrt{3} \right)^{6} < 0$

$\Longrightarrow2702-1 < 2702-\left( 2-\sqrt{3} \right)^{6} < 2702+0$

$\Longrightarrow 2701 < \left( 2+\sqrt{3} \right)^{6} < 2702$

8 0

2 years ago

Solve for N A 24 B 54 C 42 D 94.5

Nookie1986 [14]

Answer:

54 0r 42 I think

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers