Which values are solutions to the inequality below? Check all that apply. x2 25

Please help me !! Answer both correctly for a brainliest and also a thanks :)

neonofarm [45]

The firs t res is x=4
the second is D

6 0

3 years ago

How to prove that a shape is a square on the coordinate plane?

Goshia [24]

If a quadrilateral has four congruent sides and four right angles, then it's a square, and also If two consecutive sides of a rectangle are congruent, then it's a square.

Glad I could help, Have a great day!

5 0

2 years ago

Read 2 more answers

In isoceles triangle the length of a leg is 17cm, and the base is 16cm. Find the length of the altitude to the base

Elan Coil [88]

This triangle has base 16 therefore the sides must be 17cm and 17 cm

When we make a altitude it divides it into two right triangles and there is a property in which the altitude of the isoceles triangle divides the base in 2 equal halves

So the side of the right triangle will be x , 8 , 17

Using pythgoreus theorem

x²+8²=17²

x = √225

x = 15

So the altitude is 15 cm

Must click thanks and mark brainliest

6 0

2 years ago

What is the LEAST possible integer value and the greatest

vagabundo [1.1K]

Answer:

hey mate I guess ur question is incomplete......after greatest what is it? plz check

8 0

2 years ago

Read 2 more answers

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

Which values are solutions to the inequality below? Check all that apply.x2 ￼ 25

Which values are solutions to the inequality below? Check all that apply.x2 25