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2 years ago

Daniel buys a video game for $48.50. He must pay 6% sales tax. How much tax must Daniel pay?

Mathematics

2 answers:

dolphi86 [110]2 years ago

4 0

If you add 6% your answer would be $51.41 hope it helps <3

Licemer1 [7]2 years ago

4 0

He would pay 51.41 you would take 48.50 times by .06 for sales tax and you would get 2.91 then you would add 48.50+2.91 which equals 51.41

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F(t) 2t-3 F(7)=? What does f(7) equal?

myrzilka [38]

Answer:

F(7) = 11

Step-by-step explanation:

F(7) = 2(7) -3

= 14 - 3

= 11

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2 years ago

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suppose line a is parallel to line b and line b is perpendicular to line c. what relationship between line a and line c

kykrilka [37]

Answer:

a and c would also be perpendicular

Step-by-step explanation:

because a║b and b⊥c you would use the transitive property of equality.

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2 years ago

Any 10th grader solve it 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

5 0

2 years ago

Helpppppppppppppppp ill mark you brainlist write in y=mx+b form

Alex Ar [27]

Answer:

I think: 0 = 3x + b or b = -3x + 0 or x = -b + 0/3 or 3 = -b + 0/x

Sorry if you get this wrong..

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1 year ago

There are 1,295 students attending a small university.There are 714 women enrolled. What percent of the students are women. Equa

Semenov [28]

If 1295 × p = 714, then you can rearrange the equation to find the percentage.

You have to get p on its own, so you divide 1295 on both sides of the '='.

1295 × p = 714

p = 714 ÷ 1295

p = 0.55135135135...

p × 100 = actual percentage

actual percentage = 55.14%

8 0

2 years ago