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nataly862011 [7]

3 years ago

11

Phil has budgeted 120 to rent a car during his weeklong vacation. At the rate of $40 per week plus 10 cents per mile how many mi

les can he drive the car and stay within the budget

1 answer:

Elanso [62]3 years ago

4 0

800 miles 120-40=80 so he has $80 worth of miles to use. He can go 10 miles on $1 so 80*10=800

He can go 800 miles.

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Pls help i need help ASappp

Harrizon [31]

A=πr2=π·92≈254.469
your answer is 254.469

4 0

2 years ago

Read 2 more answers

Amanda is making pastry dough. She mixes of a 1/3cup of flour and of a 1/5 cup of sugar. If she wants to mix flour and sugar in

mr_godi [17]

Flour : sugar = (1/3) : (1/5) = 5 : 3 . . . . . . . . multiply the first ratio by 15 to eliminate fractions

The total of ratio units in the last representation is 5+3=8, so for a recipe of 16 cups, each ratio unit will stand for 16/8 = 2 cups.

Multiplying the ratio by 2 cups, we find Amanda will need

... 10 cups flour : 6 cups sugar

for her recipe.

4 0

3 years ago

Read 2 more answers

Please help with my math problem

REY [17]

Answer is B hope it helps

6 0

3 years ago

It takes Bruce 12 h to complete his training. It takes Oliver approximately 2 days, 2 h, and 15 min to complete

Stolb23 [73]

Answer:

Oliver completed his training work in 50.25 hours

Step-by-step explanation:

we know that

1 day= 24 hours

1 hour=60 minutes

we have

2 days, 2 h, and 15 min

<u><em>Convert days to hours</em></u>

2 days=2(24)=48 hours

<u><em>Convert minutes to hour</em></u>

15 min=15(1/60)=1/4=0.25 h

so

2 days, 2 h, and 15 min=(48)+(2)+(0.25)=50.25 h

therefore

Oliver completed his training work in 50.25 hours

3 0

3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

7 0

3 years ago

Read 2 more answers