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3 years ago

The area of a right-angléd triangle is

Mathematics

1 answer:

Brrunno [24]3 years ago

3 0

Answer:

10cm

Step-by-step explanation:

Right angle triangle area= ab/2

GO IN REVERSE

24 x2= 48

48/8=6

now you have the 2 side you need to use Pythagoras

6^2+8^2=100

square root of 100=10cm

so the hypotenuse is 10 cm

You might be interested in

If TSP and PSW are supplementary, and TSP is two times as large as mPSW, what is mTSP?

Burka [1]

Answer:

$\huge\boxed{120 \textdegree}$

Step-by-step explanation:

In order to solve for m∠TSP, we can use basic angle relationships and equations.

Let’s assume m∠PSW is $x$ °

We know that if two angles are supplementary, their angle lengths will add to 180°.

We also know that m∠TSP is twice as large as PSW. Since PSW is represented as x°, TSP will be 2x°.

We can add these two angles together and get 180°. Then you can solve for x (PSW)

$x+2x=180$
$3x=180$
$x=60$

So we know that x (PSW) is 60°. However, the question asks for the measure of TSP. We know that TSP is twice as large as PSW, so we can multiply PSW (60) by 2.

$60 \cdot 2 = 120$

So m∠TSP is 120°.

Hope this helped!

3 0

4 years ago

Calculate the value of x

vampirchik [111]

Answer:

x = 30

Step-by-step explanation:

All the angles on a point add up to 360°. Thus, you can find 2x easily.

2x = 360° - 300°

= 60

x = 60 ÷ 2

= 30

5 0

3 years ago

Read 2 more answers

Look at the box plot. What is the median? * 5 15 25 35

yaroslaw [1]

25 is the correct answer

5 0

3 years ago

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$

$t_{s} = 5.303\,h$

The draining time when only the small drain is opened is 5.303 hours.

7 0

4 years ago

PLS HELPPP !! BRAINLIEST!!

satela [25.4K]

18cm you multiple it my the figure given

6 0

4 years ago