<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B4%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20" id="TexFormula1" title=" \fra

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3 years ago

c{3}{4} + \frac{1}{2} " alt=" \frac{3}{4} + \frac{1}{2} " align="absmiddle" class="latex-formula">
please help I forgot how to do this

Mathematics

1 answer:

Gnesinka [82]3 years ago

5 0

Answer:

5/4

Step-by-step explanation:

You just need to multiply 1/2 by 2 to give you a common denominator:

3/4+2/4

From there you can add the numerators:

5/4

HTH :)

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Find the surface area of a square pyramid with a base length of 24 cm and a height of 16 cm. Explain how you solved the problem.

icang [17]

Answer:

the answer b

Step-by-step explanation:

[surface area]=[area of 4 triangles sides]+[area of square base]

[area of square base]=24*24--------> 576 cm²

[area of one triangles side]

l=slant height

l²=16²+12²-------> l²=400-----------> l=20 cm

[area of one triangles side]=24*20/2--------> 240 cm²

[area of 4 triangles sides]=4*240----------> 960 cm²

[surface area]=[960]+[576]--------> 1536 cm²

6 0

3 years ago

A water cooler can hold 50 pt of water about how many litters of water can it hold

serious [3.7K]

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3 years ago

Find the solution -18x + 3 <48

aliina [53]

Answer:

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Step-by-step explanation:

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3 0

3 years ago

Solve for the missing side to the nearest tenth. 19 X 51

telo118 [61]

Answer:

x = 30.2 units

Step-by-step explanation:

Trigonometric Ratios

The ratios of the sides of a right triangle are called trigonometric ratios.

Selecting any of the acute angles, it has an adjacent side and an opposite side. The trigonometric ratios are defined upon those sides and the hypotenuse.

The given right triangle has an angle of measure 51° and its adjacent leg has a measure of 19 units. It's required to calculate the hypotenuse of the triangle.

We use the cosine ratio to calculate x:

$\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}$

$\displaystyle \cos 51^\circ=\frac{19}{x}$

Solving for x:

$\displaystyle x=\frac{19}{\cos 51^\circ}$

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3 years ago

I wrote down 3 different positive whole numbers that add 96. the sum of any 2 is divisible by the third.

il63 [147K]

This question is really simple if you focus on this part

“The sum of any two is divisible by the third”

⟹⟹ Pairwise, the numbers must have a GCD greater than one.

Let this GCD be xx.

The first three easiest numbers I can think of are x,2xx,2x and 3x3x

Because

gcd(x,2x)=gcd(2x,3x)=gcd(3x,x)=xgcd(x,2x)=gcd(2x,3x)=gcd(3x,x)=x

The numbers sum to 9696

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⟹6x=96⟹6x=96

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The largest number is 3x=3×16=483x=3×16=48

Thanks for the A2A

4 0

3 years ago