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3 years ago

What is the length of a hypotenuse when the legs are 5 and 12

1 answer:

8 0

By using pythagoras theorem where
(Base)^2 +(perpedicular)^2 = (hypotenuse)^2
So,
(5)^2 +(12)^2 = (hypotenuse)^2
25 + 144 = (hypotenuse)^2
169 = (hypotenuse)^2
13 = hypotenuse

So, hypotenuse = 13

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Look at the triangles shown below.

Sophie [7]

Answer:

Variant c

Step-by-step explanation:

c²=4*9

c=6

You can apply phyphagor

b²=6²+9²=36+81=117

$3 \sqrt{13}$

4 0

3 years ago

What is the quotient StartFraction 15 p Superscript negative 4 Baseline q Superscript negative 6 Baseline Over negative 20 p Sup

zhannawk [14.2K]

Answer:

$- \frac{3}{4} \times \frac{p^{8} }{q^{3} }$

Step-by-step explanation:

We have to find the quotient of the following division, $\frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}$ .

Now, $\frac{15p^{-4}q^{-6} }{- 20p^{-12} q^{-3}}$

= $- \frac{3}{4} p^{[- 4 - (- 12)]} q^{[-6 - (- 3)]}$ {Since all the terms in the expression are in product form, so we can treat them separately}

{Since we know the property of exponent as $\frac{a^{b} }{a^{c} } = a^{(b - c)}$ }

= $- \frac{3}{4} p^{8} q^{-3}$

= $- \frac{3}{4} \times \frac{p^{8} }{q^{3} }$ (Answer)

{Since we know, $a^{-b} = \frac{1}{a^{b} }$ }

3 0

3 years ago

Read 2 more answers

Zia was driving his truck on the superhighway. His speed was recorded by the motorway camera between 6:00 am to 8:30 am. He cove

kodGreya [7K]

Answer:

Average speed = 100 km/h

Step-by-step explanation:

Given:

Total distance covered = 250 km

Total time taken = 8:30 am - 6:00 am = 2:30 hours = 2.5

Find:

Average speed.

Computation:

⇒ Average speed = Total distance covered / Total time taken

⇒ Average speed = 250 / 2.5

⇒ Average speed = 100 km/h

6 0

3 years ago

Plz answer this <br> 1.580 ml: 1.12 liter: 104ml

tatyana61 [14]

Answer:

Step-by-step explanation:

4 0

4 years ago

On a standard reticulocyte preparation with new methylene blue, there are 100 cells counted with blue-stained granulofilamentous

DIA [1.3K]

Answer:

0.322 × 10¹² /L

Step-by-step explanation:

Data provided in the question:

Number of cells counted with blue-stained granulofilamentous material

i.e number of RET = 100

The red blood cells count = 3.22 × 10¹² /L

Hematocrit = 30%

Now,

RET% = [ [ Number of RET ] ÷ 1000 RBCs ] × 100%

= [ 100 ÷ 1000 ] × 100%

= 0.1 × 100%

= 10%

also,

Absolute reticulocyte count = ( %RET × RBC count ) ÷ 100

= [ 10 × 3.22 × 10¹² /L ] ÷ 100

= 0.322 × 10¹² /L

8 0

3 years ago