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aleksandr82 [10.1K]

2 years ago

5

Help me with this, it's Pythagorean theorem.

2 answers:

Blizzard [7]2 years ago

6 0

Step-by-step explanation:

Let x represent the missing side length.

By Pythagoras' Theorem,

We have x² + 12² = 13².

Therefore x² = 13² - 12² = 25 and x = 5.

(x = -5 is rejected as side lengths are positive)

ziro4ka [17]2 years ago

4 0

Answer:

Let the required side of the triangle be x

By Pythagoras theorem we get,

$x = \sqrt{ {13}^{2} - {12}^{2} } \\ = \sqrt{169 - 144} \\ = \sqrt{25} \\ = \sqrt{ {5}^{2} } \\ = 5$

<h3>5 is the right answer.</h3>

You might be interested in

Use distributive law to factor the given expression: 3+21a+15b

Reil [10]

Answer:

$3(1+7a+5b)$

Step-by-step explanation:

The distributive law states that

$a\cdot (b+c)=a\cdot b+a\cdot c.$

The expression 3+21a+15b consists of three terms:

3=3·1;
21a=3·7a;
15b=3·5b.

You can see that 3 is the common factor of these three terms, thus

$3+21a+15b=3(1+7a+5b).$

6 0

3 years ago

Given: ABCD trapezoid<br> BC=1.2m, AD=1.8m<br> AB=1.5m, CD=1.2m<br> AB∩CD=K<br> Find: BK and CK

stepladder [879]

Answer:

The value of BK is 3m and the value of CK is 2.4m.

Step-by-step explanation:

Given information: ABCD trapezoid

, BC=1.2m, AD=1.8m

, AB=1.5m, CD=1.2m

, AB∩CD=K.

Using the given information draw a figure.

Two sides of a trapezoid are parallel.

Since AB∩CD=K, therefore AB and CD are not parallel, because parallel line never intersect.

$AD\parallelBC$

$\angle KBC=\angle KAD$ (Corresponding angles)

$\angle KCB=\angle KDA$ (Corresponding angles)

By AA rule of similarity

$\triangle KBC\sim \triangle KAD$

Corresponding sides of similar triangles are proportional.

$\frac{KB}{KA}=\frac{KC}{KD}=\frac{BC}{AD}$

$\frac{x}{x+1.5}=\frac{y}{y+1.2}=\frac{1.2}{1.8}$

$\frac{x}{x+1.5}=\frac{1.2}{1.8}$

$\frac{x}{x+1.5}=\frac{2}{3}$

$3x=2x+3$

$x=3$

The length of BK is 3 m.

$\frac{y}{y+1.2}=\frac{1.2}{1.8}$

$\frac{y}{y+1.2}=\frac{2}{3}$

$3y=2y+2.4$

$y=2.4$

The length of CK is 2.4 m.

5 0

2 years ago

Read 2 more answers

denis23 [38]

I think it might be A and C

8 0

3 years ago

Read 2 more answers

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago

A survey of 46 college athletes found that 24 played volleyball, while 22 played basketball. a) If we pick one athlete survey pa

PSYCHO15rus [73]

Answer:

A) $\dfrac{11}{23}$

B) $\dfrac{88}{345}$

Step-by-step explanation:

A survey of 46 college athletes found that

24 played volleyball,
22 played basketball.

A) If we pick one athlete survey participant at random, the probability they play basketball is

$P_1=\dfrac{22}{46}=\dfrac{11}{23}$

B) If we pick 2 athletes at random (without replacement),

the probability we get one volleyball player is $\dfrac{24}{46}=\dfrac{12}{23};$
the probability we get another basketball player is $\dfrac{22}{45}$ (only 45 athletes left).

Thus, the probability we get one volleyball player and one basketball player is

$P_2=\dfrac{12}{23}\cdot \dfrac{22}{45}=\dfrac{88}{345}$

5 0

3 years ago