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3 years ago

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The expression 15n + 2(3p) represents the amount Isaiah spent buying gasoline and snacks, where n represents the price of each g

allon of gasoline and p represents the cost of each snack that he bought. Which statement is true about the amount Isaiah spent?

1 answer:

Lena [83]3 years ago

3 0

Answer:

p= 2.5

Step-by-step explanation:

maybe you meant 15=2(3p)

so that would be the answer

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Decompose the longest side to find the areacof the shape. (Use the Distributive Property) the shape is 12cm x 9cm

attashe74 [19]

3(4*3)=12*9
12*9=108
3(4*3)=3*4=12
3*3=9
12*9

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3 years ago

Pls help giving brbrainlyist

chubhunter [2.5K]

The fourth choice. For example, (x+2)(x-2)>0 we'll get x>2 or x<-2 because whenever you substitute x>2 for example x = 3, the value will be higher than 0 itself.

If we substitute x = 3 (for x>2)

We'll get (3+2)(3-2)>0

5(1)>0

5>0 The statement is true.

If we substitute x = -3 (for x<-2)

We'll get (-3+2)(-3-2)>0

-1(-5)>0

5>0 The statement is true.

We don't say x>2 AND x<-2, In math, we only say x>2 OR x<-2

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3 years ago

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Any 10th grader solve it <br>for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2}-pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$=\frac{Arq^{2}-Dq^{2}r-Aqr^{2}+qr^{2}D+Apr^{2} -pDr^{2}-Ap^{2}r+p^{2}rD+Ap^{2}q-Dp^{2}q-Aq^{2}p+q^{2}pD}{pqr}$

$\neq 0$

ie., $RHS\neq 0$

Therefore $LHS\neq RHS$

ie., $\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$

Hence proved

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3 years ago

What is the value of the expression below when X=5 6x+5

Molodets [167]

Answer:

35

Step-by-step explanation:

6 x 5 = 30

30 + 5 = 35

6 0

3 years ago

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AT the local zoo, the aquarium can be seen from the Reptile Room and the Amphibian Arena. What is the total area of both rooms a

BartSMP [9]

Answer:

The total area of both rooms and the aquarium is equal to 10,932 ft²

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

the length sides of Aquarium are

(83-32) x (72-52)-----> 51 ft x 20 ft

Area reptile room=(83*72)-area of aquarium----> 83*72-51*20---> 4,956 ft²

[ total area of both rooms]=2*4,956-----> 9,912 ft²

I'm assuming that the two areas are equal

Area of aquarium=51*20=1,020 ft²

therefore

the total area of both rooms and the aquarium is equal to

9,912 ft²+1,020 ft²=10,932 ft²

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3 years ago