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3 years ago

14

Self-driving taxi service charges $5 per pickup and $0.25 per mile. We’ll Get You. There’s cabs charges $1 per pickup and $0.50

per mile. At how many miles will the two companies charge you the same amount?

1 answer:

hoa [83]3 years ago

6 0

Let y = total cost of the taxi service.

let x = amount of miles

We need to create a system of equations to solve this problem.

Taxi company 1: y = $0.25x + $5

Taxi company 2: y = $0.50x + $1

Set the equations equal to each other

$0.25x + $5 = $0.50x +$1.

Subtract $0.25x from both sides and subtract $1 from both sides.

$4 = $0.25x

Divide both sides by $0.25.

16 miles = x

The companies will charge the same amount of money at 16 miles

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A soap factory packs 7 bars of soap into each family pack. How many family packs can be made with 4,900 bars of soap?

m_a_m_a [10]

Answer:

700 family packs

Step-by-step explanation:

In each family pack, there are 7 bars of soap. To find the number of family packs that can be made with 4900 bars of soap, we need to divide 4900 into groups of 7.

4900÷7

=700

Therefore, 700 family packs can be made with 4900 bars of soap.

I hope this helps!

5 0

3 years ago

(Prove) The angle subtended by an arc at the center is double the angle subtended by it at any

Pavlova-9 [17]

Answer:

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Step-by-step explanation:

Let us consider the image attached.

Center of circle be O.

Arc AB subtends the angle $\angle APB$ on the circle and $\angle AOB$ on the center of the circle.

To prove:

$\angle AOB = 2 \times \angle APB$

Proof:

In $\triangle PAO$ : AO and PO are radius of the circles so AO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, $\angle PAO = \angle OPA$

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

$\angle AOQ = \angle PAO + \angle OPA=2 \times \angle APO .... (1)$

Similarly,

In $\triangle PBO$ : BO and PO are radius of the circles so BO = PO

And angles opposite to equal sides of a triangle are also equal in a triangle.

So, $\angle PBO = \angle OPB$

Using external angle property, that external angle is equal to sum of two opposite internal angles of a triangle.

$\angle BOQ = \angle PBO + \angle OPB=2 \times \angle BPO .... (2)$

Now, we can see that:

$\angle AOB = \angle AOQ+\angle BOQ$

Using equations (1) and (2):

$\angle AOB = 2\angle APO+2\angle BPO\\\angle AOB = 2(\angle APO+\angle BPO)\\\bold{\angle AOB = 2(\angle APB)}$

Hence, proved.

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3 years ago

1,000,000 meters equal 1 megameter.<br><br> True<br><br> False

a_sh-v [17]

The answer is TRUE.

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2 years ago

Seven times the sum of twice a number and 16

Brut [27]

Answer:

7(2x+16)

Step-by-step explanation:

first, we look at the part where it says the sum of twice a number and 16

let's represent the number by using x. since it says twice a number, that would mean 2x.

when it says the sum of twice a number and 16, this means that we have to add them, so we would get 2x+16.

finally, in the beginning it says seven times the sum of twice a number and x, which would mean that we would have to multiple 2x+16 by 7.

answer:

7(2x+16)

6 0

3 years ago

Read 2 more answers

HELP NEEDED. 37 POINTS<br>I just need the answers

Juli2301 [7.4K]

Answer:

Part 1) $P=[2\sqrt{29}+\sqrt{18}]\ units$ or $P=15.01\ units$

Part 2) $P=2[\sqrt{20}+\sqrt{45}]\ units$ or $P=22.36\ units$

Part 3) $P=4[\sqrt{13}]\ units$ or $P=14.42\ units$

Part 4) $P=[19+\sqrt{17}]\ units$ or $P=23.12\ units$

Part 5) $P=2[\sqrt{17}+\sqrt{68}]\ units$ or $P=24.74\ units$

Part 6) $A=36\ units^{2}$

Part 7) $A=20\ units^{2}$

Part 8) $A=16\ units^{2}$

Part 9) $A=10.5\ units^{2}$

Part 10) $A=6.05\ units^{2}$

Step-by-step explanation:

we know that

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Part 1) we have the triangle ABC

$A(0,3),B(5,1),C(2,-2)$

step 1

Find the distance AB

$A(0,3),B(5,1)$

substitute in the formula

$AB=\sqrt{(1-3)^{2}+(5-0)^{2}}$

$AB=\sqrt{(-2)^{2}+(5)^{2}}$

$AB=\sqrt{29}\ units$

step 2

Find the distance BC

$B(5,1),C(2,-2)$

substitute in the formula

$BC=\sqrt{(-2-1)^{2}+(2-5)^{2}}$

$BC=\sqrt{(-3)^{2}+(-3)^{2}}$

$BC=\sqrt{18}\ units$

step 3

Find the distance AC

$A(0,3),C(2,-2)$

substitute in the formula

$AC=\sqrt{(-2-3)^{2}+(2-0)^{2}}$

$AC=\sqrt{(-5)^{2}+(2)^{2}}$

$AC=\sqrt{29}\ units$

step 4

Find the perimeter

The perimeter is equal to

$P=AB+BC+AC$

substitute

$P=[\sqrt{29}+\sqrt{18}+\sqrt{29}]\ units$

$P=[2\sqrt{29}+\sqrt{18}]\ units$

or

$P=15.01\ units$

Part 2) we have the rectangle ABCD

$A(-4,-4),B(-2,0),C(4,-3),D(2,-7)$

Remember that in a rectangle opposite sides are congruent

step 1

Find the distance AB

$A(-4,-4),B(-2,0)$

substitute in the formula

$AB=\sqrt{(0+4)^{2}+(-2+4)^{2}}$

$AB=\sqrt{(4)^{2}+(2)^{2}}$

$AB=\sqrt{20}\ units$

step 2

Find the distance BC

$B(-2,0),C(4,-3)$

substitute in the formula

$BC=\sqrt{(-3-0)^{2}+(4+2)^{2}}$

$BC=\sqrt{(-3)^{2}+(6)^{2}}$

$BC=\sqrt{45}\ units$

step 3

Find the perimeter

The perimeter is equal to

$P=2[AB+BC]$

substitute

$P=2[\sqrt{20}+\sqrt{45}]\ units$

or

$P=22.36\ units$

Part 3) we have the rhombus ABCD

$A(-3,3),B(0,5),C(3,3),D(0,1)$

Remember that in a rhombus all sides are congruent

step 1

Find the distance AB

$A(-3,3),B(0,5)$

substitute in the formula

$AB=\sqrt{(5-3)^{2}+(0+3)^{2}}$

$AB=\sqrt{(2)^{2}+(3)^{2}}$

$AB=\sqrt{13}\ units$

step 2

Find the perimeter

The perimeter is equal to

$P=4[AB]$

substitute

$P=4[\sqrt{13}]\ units$

or

$P=14.42\ units$

Part 4) we have the quadrilateral ABCD

$A(-2,-3),B(1,1),C(7,1),D(6,-3)$

step 1

Find the distance AB

$A(-2,-3),B(1,1)$

substitute in the formula

$AB=\sqrt{(1+3)^{2}+(1+2)^{2}}$

$AB=\sqrt{(4)^{2}+(3)^{2}}$

$AB=5\ units$

step 2

Find the distance BC

$B(1,1),C(7,1)$

substitute in the formula

$BC=\sqrt{(1-1)^{2}+(7-1)^{2}}$

$BC=\sqrt{(0)^{2}+(6)^{2}}$

$BC=6\ units$

step 3

Find the distance CD

$C(7,1),D(6,-3)$

substitute in the formula

$CD=\sqrt{(-3-1)^{2}+(6-7)^{2}}$

$CD=\sqrt{(-4)^{2}+(-1)^{2}}$

$CD=\sqrt{17}\ units$

step 4

Find the distance AD

$A(-2,-3),D(6,-3)$

substitute in the formula

$AD=\sqrt{(-3+3)^{2}+(6+2)^{2}}$

$AD=\sqrt{(0)^{2}+(8)^{2}}$

$AD=8\ units$

step 5

Find the perimeter

The perimeter is equal to

$P=AB+BC+CD+AD$

substitute

$P=[5+6+\sqrt{17}+8]\ units$

$P=[19+\sqrt{17}]\ units$

or

$P=23.12\ units$

Part 5) we have the quadrilateral ABCD

$A(-1,5),B(3,6),C(5,-2),D(1,-3)$

step 1

Find the distance AB

$A(-1,5),B(3,6)$

substitute in the formula

$AB=\sqrt{(6-5)^{2}+(3+1)^{2}}$

$AB=\sqrt{(1)^{2}+(4)^{2}}$

$AB=\sqrt{17}\ units$

step 2

Find the distance BC

$B(3,6),C(5,-2)$

substitute in the formula

$BC=\sqrt{(-2-6)^{2}+(5-3)^{2}}$

$BC=\sqrt{(-8)^{2}+(2)^{2}}$

$BC=\sqrt{68}\ units$

step 3

Find the distance CD

$C(5,-2),D(1,-3)$

substitute in the formula

$CD=\sqrt{(-3+2)^{2}+(1-5)^{2}}$

$CD=\sqrt{(-1)^{2}+(-4)^{2}}$

$CD=\sqrt{17}\ units$

step 4

Find the distance AD

$A(-1,5),D(1,-3)$

substitute in the formula

$AD=\sqrt{(-3-5)^{2}+(1+1)^{2}}$

$AD=\sqrt{(-8)^{2}+(2)^{2}}$

$AD=\sqrt{68}\ units$

step 5

Find the perimeter

The perimeter is equal to

$P=\sqrt{17}+\sqrt{68}+\sqrt{17}+\sqrt{68}$

substitute

$P=2[\sqrt{17}+\sqrt{68}]\ units$

or

$P=24.74\ units$

<h3>The complete answer in the attached file</h3>

8 0

3 years ago