All categories

3 years ago

two sides of a triangle measure 20 cm and 30 cm. which of the following could be the measure of the third side.

Mathematics

1 answer:

Alexandra [31]3 years ago

6 0

Answer:

the answer is b 6 cm your welcome

You might be interested in

Thea and Eleanor are reading the same book. The graph shows the number of words Thea reads over time. Eleanor reads 225 words pe

Lapatulllka [165]

The slope is the rate
words per minute
we can do
540 words in 3 minutes
divide by 3
180 words in 1 minute

thea=180wpm
elanor=225wpm

elanor>thea

and 225*2=450

answers are
Thea reads 180 words per minute.

Eleanor reads 450 words every 2 minutes.

After 1 hour of reading, Eleanor reads more words than Thea.

6 0

3 years ago

Read 2 more answers

Biologists are studying a new bacterium. They create a culture with 100 of the bacteria and anticipate that the number of bacter

Ksju [112]

Answer:

3000

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

I'm having trouble figuring out how I could go about this. Can someone work it out then explain so that I know how to do this ne

skad [1K]

Answer:

RT-ST=QS-ST, because subtracting the same quantity from two lines that have been stated to be equal.

Therefore RS=TQ

Angle R=Angle Q, because it is an isosceles triangle

Triangle AR

S is congruent to ATQ.

AT=AS

Because TAS is isosceles, angles 5 and 6 are equal.

Therefore, angles 4 and 7 are equal, because they are supplementary angles of the same angle.

And angles 1 and 3 are equal, because the other two angles in the triangle are equal.

The triangles RAT and QAS are congruent with SAS.

Step-by-step explanation: Can u gimme brain plz!

6 0

3 years ago

Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join

zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx = 1$

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y) and replacing a and b with their respective values, we have

$\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx = 1$

Since c is a constant, we can bring it out of the integral sign; to give us

$c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx = 1$

Open the bracket

$c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx = 1$

Integrate with respect to y

$c\int\limits^3_0 {\frac{xy^{2}}{2} +\frac{xy^{3}}{3} } \, dx (0,3}) = 1$

Substitute 0 and 3 for y

$c\int\limits^3_0 {(\frac{x* 3^{2}}{2} +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2} +\frac{x * 0^{3}}{3})} \, dx = 1$

$c\int\limits^3_0 {(\frac{x* 9}{2} +\frac{x * 27}{3} ) - (0 +0) \, dx = 1$

$c\int\limits^3_0 {(\frac{9x}{2} +\frac{27x}{3} ) \, dx = 1$

Add fraction

$c\int\limits^3_0 {(\frac{27x + 54x}{6}) \, dx = 1$

$c\int\limits^3_0 {\frac{81x}{6} \, dx = 1$

Rewrite;

$c\int\limits^3_0 (81x * \frac{1}{6}) \, dx = 1$

The $\frac{1}{6}$ is a constant, so it can be removed from the integral sign to give

$c * \frac{1}{6}\int\limits^3_0 (81x ) \, dx = 1$

$\frac{c}{6}\int\limits^3_0 (81x ) \, dx = 1$

Integrate with respect to x

$\frac{c}{6} * \frac{81x^{2}}{2} (0,3) = 1$

Substitute 0 and 3 for x

$\frac{c}{6} * \frac{81 * 3^{2} - 81 * 0^{2}}{2} = 1$

$\frac{c}{6} * \frac{81 * 9 - 0}{2} = 1$

$\frac{c}{6} * \frac{729}{2} = 1$

$\frac{729c}{12} = 1$

Multiply both sides by $\frac{12}{729}$

$c = \frac{12}{729}$

$c = 0.0165 (Approximately)$

8 0

3 years ago

(x-1)(x-3)(x+5)(x+7)=297

mihalych1998 [28]

First simplify the expression into polynomial form,

$(x-1)(x-3)(x+5)(x+7)=297$

$x^4+8x^3-10x^2-104x+105=297$

$x^4+8x^3-10x^2-104x-192=0$

Now factor into,

$(x-4)(x+8)(x^2+4x+6)=0$

Which means the solutions are,

$x-4=0\implies\boxed{x_1=4}$

$x+8=0\implies\boxed{x_2=-8}$

and then two complex solutions because determinant of the third factor $D\lt0$ ,

$x^2+4x+6=0$

$x^2+4x+4=-2$

$(x+2)^2=-2\implies\boxed{x_3=i\sqrt{2}-2},\boxed{x_4=-i\sqrt{2}-2}$

Hope this helps :)

5 0

3 years ago

Read 2 more answers