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3 years ago

A bag contains 4 red marbles, 3 blue marbles and 3 yellow marbles. Match each statement with the correct value.

Mathematics

2 answers:

MAXImum [283]3 years ago

8 0

Answer: thank me :))

Step-by-step explanation:

tia_tia [17]3 years ago

4 0

I’m having a hard time understanding your question

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How many cubic feet of gas will a rectangular gas tank hold if it measures 15 ft x 4 ft x 6 ft?

AlladinOne [14]

15 ft X 4 ft = 60 ft
60 ft X 6 ft = 360 cubic ft

7 0

3 years ago

What is the value of the expression 18-2 times ( 4 +3)

Snowcat [4.5K]

Answer:

1 2/7

Step-by-step explanation:

18 - 2x(4+3) =

combine like terms in the parentheses

18 - 2x(7) =

Multiply

18 - 14x (This is the value of the expression)

We can solve for x by setting it equal to zero

18 - 14x = 0

-14x = -18

x = -18/-14

x = 1 2/7

5 0

2 years ago

A football is kicked toward the goal. the height of the ball is modeled by the function h(t) = −16t2 64t, where t equals the tim

xenn [34]

Answer:

axis of symmetry : x = 4

Step-by-step explanation:

The x -coordinate of the vertex is the equation of the axis of symmetry of the parabola.

The vertex of h(t) = -16t^2 + 64x is ( 4 , 0 )

The vertex is the highest point of the graph so it represents that after four seconds the ball is at its highest height when the football is being kicked towards the goal.

6 0

1 year ago

Calculate the slope of the graph below. Write your answer into the blank, m=

Mekhanik [1.2K]

Answer: 2

Step-by-step explanation:

Slope=rise/run

In this case

Rise=2

Run=1

2/1=2

5 0

3 years ago

Read 2 more answers

Using Cramer's rule to solve linear systems.

Rzqust [24]

Answer: Last Option

$x=2,\ y=-5$

Step-by-step explanation:

Cramer's rule says that given a system of equations of two variables x and y then:

$x =\frac{Det(A_X)}{Det(A)}$

$y =\frac{Det(A_Y)}{Det(A)}$

For this problem we know that:

$Det(A) = |A|=\left|\begin{array}{ccc}4&-6\\8&-2\\\end{array}\right|$

Solving we have:

$|A|= 4*(-2) -(-6)*8\\\\|A|=40$

$Det(A_X) = |A_X|=\left|\begin{array}{ccc}38&-6\\26&-2\\\end{array}\right|$

Solving we have:

$|A_X|=38*(-2) - (-6)*26\\\\|A_X|=80$

$Det(A_Y) = |A_Y|=\left|\begin{array}{ccc}4&38\\8&26\\\end{array}\right|$

Solving we have:

$|A_Y|=4*(26) - (38)*8\\\\|A_Y|=-200$

Finally

$x =\frac{|A_X|}{|A|} = \frac{80}{40}\\\\x=2$

$y =\frac{|A_Y|}{|A|} = \frac{-200}{40}\\\\y=-5$

4 0

3 years ago