LOTS OF POINTS!!!<br> SOLVE ANd EXPLAIN PROBLEM 8

PLZ HELP I AM GONNA FAIL JUST EXPLAIN WHAT TO DO

Alexxandr [17]

A number cube is a die (singular for dice). List all of the possible outcomes of a die (list numbers of 1 to 6), and next to those numbers, determine if they are even or odd.

8 0

3 years ago

Write the point-slope form of the equation of the line passing through the points (-5, 6) and (0, 1).

andrey2020 [161]

Formula of the slope of a linear function:

m = (y₂ - y₁)/(x₂ - x₁)

m= (-1 - 6)/[0- (-5)]

m= -7/5 This is the slope requested

7 0

4 years ago

A bag contains five red marbles, four blue marbles, and four yellow marbles. You randomly pick a marble. What is the probability

Vikentia [17]

n(s)= 5+4+4=13

n(e)= red + blue

=5+4

=9

so, p(e) = n(e)/n(s)

p(e)= 9/13

=0.69

it's your answer

it will help.

7 0

3 years ago

Given the right triangle shown below.if Cos A =15/17 and tan A=8/15 which of the following represents the length of BC

otez555 [7]

Answer:

The Figure for Right triangle is below,

Therefore , 15 unit length represents BC.

Step-by-step explanation:

Given:

Consider a right triangle ABC, Such that

$\cos A=\dfrac{15}{17}\\\\\tan A=\dfrac{8}{15}$

To Find:

BC = ?

Solution:

In Right Angle Triangle ABC, Cosine and Tangent identity

$\cos A = \dfrac{\textrm{side adjacent to angle C}}{Hypotenuse}\\$

$\tan A= \dfrac{\textrm{side opposite to angle A}}{\textrm{side adjacent to angle A}}$

BUT,

$\cos A=\dfrac{15}{17}\\\\\tan A=\dfrac{8}{15}$ ....Given

On Comparing,

Adjacent side to angle A = AB = 15

Opposite side to angle A = BC = 8

Hypotenuse = AC =17

Also Pythagoras theorem is Satisfies,

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

$(\textrm{Hypotenuse})^{2} = 17^{2}=289$

$(\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}=15^{2}+8^{2}=289$

The Figure for Right triangle is below,

Therefore , 15 unit length represents BC.

7 0

3 years ago

A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an uppe

Ganezh [65]

Answer:

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

$Volume=\frac{1}{2} (b_{1}+4+2x)h X H$