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3 years ago

I need help with this one please

Mathematics

2 answers:

Natasha_Volkova [10]3 years ago

8 0

6
6
............................

Nesterboy [21]3 years ago

5 0

Answer:

Step-by-step explanation:

2 x 3 = 6

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I BEG OF YOU HELPPPP Twice last month, Judy Carter rented a car in Fresno, California, and traveled around the Southwest on bu

sineoko [7]

Answer:

the daily fee =33 dollars

and the mileage charge.=0.35

Step-by-step explanation:

let d: be daily fee and m for mileage

cost of rental =(d*number of days)+ (m*number of mileage)

her first trip: 4d+440m=286

her second trip: 3d+190m=165.5

solve by addition and elimination

4d+440m=286 ⇒ multiply by 3 ⇒12d +1320m=(3)286

3d+190m=165.5⇒ multiply by 4⇒12d+190(4)m=4(165.5)

12d+1320m=858

12d+760m=662

subtract two equation to eliminate d

12d+1320m-12d-760m=858-662

560m=196

m=7/20=0.35 for on mileage

d: 4d+440m=286

4d=286-440(0.35)

d=(286-154)/4 33 dollars

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4 years ago

IM GOING TO GIVE 251 POINTS TO WHOEVER ANSWER THIS PROBLEM.

Mariulka [41]

Answer:

(x^87*(x^2+4))/54

Step-by-step explanation:

4 0

2 years ago

18=x+5x im confused on this problem

DaniilM [7]

18= +5+x
x=3
If you do 3•5, you get 15.
Then do 3+15, you get 18.

3 0

4 years ago

Read 2 more answers

Solve: (-5/6) × (9/20) + (-5/6) × 7/25 = ?

jonny [76]

$\bf \underline{★ How \:to\: do -} \\$

Here, we are given with four fractions to multiply two of them and to add two of them. If we add them directly by taking the LCM and adding them is not a similar way. We can clearly observe that in those four fractions, we have two fractions as common i.e, we have two fractions as same. If we have two fractions or numbers as same, we can solve the sum by an other concept called as distributive property. In this property, we multiply the common fraction with the sum of other two fractions. This concept can also be done with fractions as well as integers. So, let's solve!!

$\:$

$\bf \underline{➤ Solution-} \\$

${\tt \leadsto \dfrac{(-5)}{6} \times \dfrac{9}{20} + \dfrac{(-5)}{6} + \dfrac{7}{25}}$

Group the non-common fractions in bracket.

${\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{9}{20} + \dfrac{7}{25} \bigg)}$

First we should solve the numbers in bracket.

LCM of 20 and 25 is 100.

${\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{9 \times 5}{20 \times 5} + \dfrac{7 \times 4}{25 \times 4} \bigg)}$

Multiply the numerators and denominators in the bracket.

${\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{45}{100} + \dfrac{28}{100} \bigg)}$

Now, write both numerators in bracket with a common denominator.

${\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{45 + 28}{100} \bigg)}$

Now, add the numerators in bracket.

${\tt \leadsto \dfrac{(-5)}{6} \times \bigg( \dfrac{73}{100} \bigg)}$

Write the numerator and denominator in lowest form by cancellation method.

${\tt \leadsto \dfrac{\cancel{(-5)} \times 73}{6 \times \cancel{100}} = \dfrac{(-1) \times 73}{6 \times 20}}$

Now, multiply the numerators and denominators.

${\tt \leadsto \dfrac{(-73)}{120}}$

$\:$

${\red{\underline{\boxed{\bf So, \: the \: answer \: obtained \: is \: \: \dfrac{(-73)}{120}}}}}$

3 0

3 years ago

35 99/100 in simplest form

sergey [27]

$35 \frac{99}{100}$ is already in simplest form...so it can't be simplified any further because;
99 doesn't go into 100 evenly nor could be divided by a same number. :D

4 0

3 years ago