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3 years ago

100 points for answer. please help

Mathematics

2 answers:

Anna35 [415]3 years ago

7 0

Answer:

19.09

Step-by-step explanation:

MrMuchimi3 years ago

6 0

Answer:

19.09

Step-by-step explanation:

4x÷3+8x÷3

4x+8x÷3×3

12x²÷6

x²=6/12

x²=3/6

3/6x²

√3/6

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How can you solve for x in the proportion StartFraction 7 over 8 EndFraction = StartFraction x over 24 EndFraction? Set the sum

dusya [7]

Answer: the last one

Step-by-step explanation: because 7 times 24 equals 168 and x would be 21 so 21 times 8 equals 168.

5 0

3 years ago

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7. Bill and Steve both owe their mother money. Bill owes his mother $300 and plans

Free_Kalibri [48]

Answer:

5 weeks

$175

Step-by-step explanation:

Bill owes his mother $300 and plans to pay her $25 every week.

Therefore, after w weeks he has to give her mother more C = 300 - 25w, amount of money.

Steve owes his mother $550 and plans to pay her $75 every week.

Therefore, after w weeks he has to give her mother more C = 550 - 75w, amount of money.

If after w weeks they both will owe their mother the same amount of money, then

300 - 25w = 550 - 75w

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Now, the amount of money will be $(300 - 25 × 5) = $175. (Answer)

4 0

3 years ago

-0.9+28 ---- 6 help me solve the task im a little lazy

Anvisha [2.4K]

Answer:

around 4.52.

Fraction form: $4\frac{52}{100}$

Step-by-step explanation:

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7 0

3 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

2 years ago

Solve the system using a matrix. NEED HELP ASAP!!:):)

icang [17]

1,-4 sorry if wrong :(

7 0

2 years ago

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