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3 years ago

Will mark Brainliest!!!!!!!!

1 answer:

4 0

F(x)=25x+100To find the second month substitute x with 2. f(2)=25(2)+100month 2= $150 Do the same for month ten but substitute x with 10 this time.f(10)=25(10)+100month 10= 350 The average rate of change is $25.

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A newspaper and three hot chocolates cost $7. Two newspapers and two hot chocolates cost $6. How much does one hot chocolate cos

pochemuha

Answer:

the cost of one hot chocolate is $2.

Step-by-step explanation:

Let x be the hot chocolate cost and y be the newspaper cost.

Given:

A newspaper and three hot chocolates cost $7

So, the equation is

$3x+y=7$ --------(1)

Two newspapers and two hot chocolates cost $6

So, the second equation is.

$2x+2y=6$ ---------(2)

Find one hot chocolate cost?

Solve equation 1 for y.

$3x+y=7$

$y=7-3x$

Put y value in equation 2

$2x+2(7-3x)=6$

$2x+14-6x=6$

$14-4x=6$

$-4x=6-14$

$-4x=-8$

negative sign is cancelled both side.

$x=\frac{8}{4}$

$x=2$

Therefore, the cost of one hot chocolate is $2.

3 0

3 years ago

You're given a side length of 7 centimeters. How many equilateral triangles can you construct using this information?

leva [86]

Answer:

One

Step-by-step explanation:

An equilateral triangle has equal side lengths.

With the given information, you can construct one equilateral triangle, with the sides measuring 7cm each.

The sides of the equilateral triangle can't be mixed up, they must stay the same.

Hope this helps.

7 0

3 years ago

PLEASE HELP! ( no random links ) correct answers please. Find median

BARSIC [14]

Answer:

:) hope it helped!

3 0

3 years ago

The legs of an isosceles right triangle are 5 inches long. What is the length of the hypotenuse of the triangle to the nearest i

In-s [12.5K]

The legs are 5 inches long.

Use the Pythagorean theorem:
$5^2+5^2=c^2 \\
25+25=c^2 \\
50=c^2 \\
c=\sqrt{50} \\
c=\sqrt{25 \times 2} \\
c=5 \sqrt{2} \\
c \approx 7$

The length of the hypotenuse of the triangle is approximately 7 inches long.

6 0

3 years ago

Read 2 more answers

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 2t, 1 + 6t, 1 + 14t . Find the points at which their p

GuDViN [60]

Answer:

A) points at which paths intersect : (1,1,1) ; (2,4,8)

B) DNE

Step-by-step explanation:

A) To find the points in which the particle paths intersect, it is necessary to find the values of t for which the three components of both vectors are equal:

$t_1=1+2t_2\\\\t_1^2=1+6t_2\\\\t_1^3=1+14t_2$

you replace t1 from the first equation in the second equation:

$(1+2t_2)^2=1+6t_2\\\\1+4t_2+4t_2^2=1+6t_2\\\\4t_2^2-2t_2=0\\\\t_2(2t_2-1)=0\\\\t_2=0\\\\t_2=\frac{1}{2}$

Then, for t2 = 0 and t2=1/2 you obtain for t1:

$t_1=1+2(0)=1\\\\t_1=1+2(\frac{1}{2})=2$

Hence, for t1=1 and t2=0 the paths intersect. Furthermore, for t1=2 and t2=1/2 the paths also intersect.

The points at which the paths intersect are:

$r_1(1)=(1,1,1)=r_2(0)=(1,1,1)\\\\r_1(2)=(2,4,8)=r_2(\frac{1}{2})=(2,4,8)$

B) You have the following two trajectories of two independent particles:

$r_1(t)=(t,t^2,t^3)\\\\r_2(t)=(1+2t,1+6t,1+14t)$

To find the time in which the particles collide, it is necessary that both particles are in the same position on the same time. That is, each component of the vectors must coincide:

$t=1+2t\\\\t^2=1+6t\\\\t^3=1+14t$

From the first equation you have:

$t=1+2t\\\\t=-1$

This values does not have a physical meaning, then, the particle do not collide

answer: DNE

5 0

3 years ago