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Tema [17]
3 years ago
13

Help pls I need this done :c

Mathematics
2 answers:
kolbaska11 [484]3 years ago
5 0

He is wrong.

The factors of 91 are: 1, 7, 13, and 91.

He can make 13 teams of 7 or 7 teams of 13.

statuscvo [17]3 years ago
3 0

Answer:

He's wrong.

Step-by-step explanation:

He could make 7 teams of 13 people each because

13 x 7 = 91  &  91 ÷ 7 = 13

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Pleaseeee help! I'm really confused.
seropon [69]
He will be making a right triangle with the legs being 24 x 45. The diagonal will be the hypotenuse of the triangle. To find the amount of fencing we need the perimeter of the triangle.
Use Pythagorean theorem to find the length of the hypotenuse.
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576 + 2025 = c^2
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sqrt 2601 = c
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Adding the three lengths the total fencing needed is
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3 years ago
Please help quickly I don’t understand
bekas [8.4K]

Answer:

15y^5

Step-by-step explanation:

Formula for the area of a rectangle is length times width.

so in this case, it would be 15y^5!! Trust me I just did this!

5 0
2 years ago
Read 2 more answers
Anya buys some avocados that cost $1.50 each. What is the total cost of her
Aneli [31]
Answer C
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5 0
2 years ago
Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!
il63 [147K]

Answer:

x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

Step-by-step explanation:

w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}

6 0
3 years ago
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