!I will mark you BRAINLIEST!Can someone please help me with these questions

Mathematics

2 answers:

Alenkinab [10]3 years ago

5 0

35. It doesn’t specify what type of triangle but if it’s referring to a right triangle it would be 10 if the missing side is the hypotenuse.
If not then it can also be 6 or 8 if it’s an isosceles triangle. I seen a lot of your questions and they don’t seem to have a good clarification of what they mean exactly and I’m an honors pre cal student

alukav5142 [94]3 years ago

4 0

Answer: 35

Step-by-step explanation:

You might be interested in

Seventh graders collected some gold sand on a class field trip.

satela [25.4K]

Answer:

Class 7A collected 4.8 ounces of pure gold.

Step-by-step explanation:

Key skills required are: Percentages, Multiplication

We only need the information about Class 7A. They collected 40 oz that contains 12% gold.
In other words, this means that 12% of that 40 oz gold sand is pure gold.

Here we have to do 12% x 40 to find the number of oz of pure gold. We first have to convert 12% into a decimal. Divide it by a 100 (or move the decimal point 2 places to the left) and you will get 0.12.

Do 0.12 x 40 and you will get 4.8

Therefore, there are 4.8 oz of pure gold in Class 7A's gold sand

3 0

3 years ago

Please answer correctly !!!!!!!!! Will Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

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

Step-by-step explanation:

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

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