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3 years ago

How many 3/4 are in 12 2/4

Mathematics

1 answer:

arlik [135]3 years ago

8 0

Answer: 16 2/3

Step-by-step explanation:

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25x -5 + 32 = (when x = 3)

Ugo [173]

Answer:

102

Step-by-step explanation:

25 x 3 - 5 + 32

75 - 5 + 32

70 + 32

=102

4 0

3 years ago

Read 2 more answers

Classify the following triangle. Check all that apply.

krok68 [10]

Answer:

I will list main feature for each property and you can choose. You can comment if you want to check or have questions

Step-by-step explanation:

Isosceles - two sides are equal

Equilateral - all sides are equal

Obtuse - one angle is greater than 90 degrees

Right - one angle is equal to 90 degrees

Scalene - none of the sides is equal

Acute - all angles are less than 90 degrees

Go down the list and see if it's true or false about the given triangle

5 0

2 years ago

Use the slope-intercept method to graph the equation y=2x-4. Is (3,2) a solution?

Andru [333]

Yes, (3, 2) is a solution to y = 2x - 4. (:

8 0

3 years ago

Read 2 more answers

I’m having a hard time understanding substitution. Please help!

Vikentia [17]

$b_1=\frac{2A}{h}-b_2$

Step-by-step explanation:

Given expression is;

$A=\frac{h(b_1+b_2)}{2}$

Solving for $b_1$

Multiplying both sides by 2

$2*A=\frac{h(b_1+b_2)}{2}*2\\2A=h(b_1+b_2)$

Dividing both sides by h

$\frac{2A}{h}=\frac{h(b_1+b_2)}{h}\\\frac{2A}{h}=b_1+b_2$

Subtracting $b_2$ from both sides

$\frac{2A}{h}-b_2=b_1+b_2-b_2\\\frac{2A}{h}-b_2=b_1\\b_1=\frac{2A}{h}-b_2$

$b_1=\frac{2A}{h}-b_2$

Keywords: division, subtraction

Learn more about subtraction at:

#LearnwithBrainly

3 0

3 years ago

(10.02)

melisa1 [442]

Answer:

$sin O=\dfrac{3\sqrt{13}}{13}\\cos O=\dfrac{2\sqrt{13}}{13}\\tan O=\dfrac{3}{2}$

Step-by-step explanation:

If the point (2,3) is on the terminal side of an angle in standard position.

Adjacent of O, x=2,

Opposite of O, y=3

Next, we determine the hypotenuse, r using Pythagoras Theorem.

$Hypotenuse =\sqrt{Opposite^2+Adjacent^2} \\r=\sqrt{3^2+2^2} \\r=\sqrt{13}$

Therefore:

$sin O=\dfrac{Opposite}{Hypotenuse} \\sin O=\dfrac{3}{\sqrt{13}} \\$Rationalizing\\sin O=\dfrac{3\sqrt{13}}{13}$

$cos O=\dfrac{Adjacent}{Hypotenuse} \\cos O=\dfrac{2}{\sqrt{13}} \\$Rationalizing\\cos O=\dfrac{2\sqrt{13}}{13}$

$Tan O=\dfrac{Opposite}{Adjacent} \\tan O=\dfrac{3}{2}$

4 0

3 years ago