All categories

3 years ago

× - 10 = 4. That's the full question

Mathematics

1 answer:

azamat3 years ago

5 0

Answer:

x-10=4

×=4+10

×=14

..............

You might be interested in

BRAINLIESTT ASAP! PLEASE HELP ME :)

Anvisha [2.4K]

Step-by-step explanation:

$\displaystyle y = Asec(Bx - C) + D$

According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:

$\displaystyle Phase\:[Horisontal]\:Shift → \frac{0}{4} = 0 \\ Period → \frac{2}{4}π = \frac{π}{2}$

Therefore we have our answer.

Extended Information on the trigonometric function

$\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|$

NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the midline where the amplitude is. Moreover, ALL tangent, secant, cosecant, and cotangent functions have NO AMPLITUDE.

I am joyous to assist you anytime.

3 0

3 years ago

Jared buys 3 pounds of chicken at $3.15 per pound and 1 pound of broccoli at $1.40 per pound. How much will it cost to make dinn

zaharov [31]

Answer:

Cost of dinner: $10.85

Money to add: $1.40

Step-by-step explanation:

1. Calculate the cost of the chicken and broccoli: 3(3.15)+1(1.40)=$10.85

2. Subtract Jared's money from the cost of dinner: $10.85-$9.45=$1.40

4 0

3 years ago

Please help I don’t understand please!!!

Vaselesa [24]

Answer:

y>5.5

Step-by-step explanation:

4y+3>2y+14

Subtract 2y from both sides.

2y+3>14

Subtract 3 from both sides.

2y>11

Divide 2 by both sides.

y>5.5

-hope it helps

7 0

2 years ago

Read 2 more answers

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

3 years ago

Let n1equals50, Upper X 1equals30, n2equals50, and Upper X 2equals10. Complete parts (a) and (b) below. a. At the 0.05 leve

Viefleur [7K]

Answer:

Null hypothesis: $p_{1} = p_{2}$

Alternative hypothesis: $p_{1} \neq p_{2}$

$z=\frac{0.6-0.2}{\sqrt{0.4(1-0.4)(\frac{1}{50}+\frac{1}{50})}}=4.082$

$p_v =2*P(Z>4.082)=4.46x10^{-5}$

So the p value is a very low value and using any significance level for example $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion 1 is significantly different from proportion 2.