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2 years ago

Can you help me please

Mathematics

2 answers:

Salsk061 [2.6K]2 years ago

7 0

It’s 120
Hope this helps

Alina [70]2 years ago

3 0

Answer:

120 in²

Hope this helps!

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1) Two apples weight________ kilograms.

ratelena [41]

Answer:

12, 72, 180

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

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

5 0

2 years ago

What is (-10,8) in point- slope form?

nata0808 [166]

There can't be a point-slope form until you have TWO points.
A single point doesn't have any other form.
There are an infinite number of lines that all go through a single point.
Every one of them has a different slope, and a different point-slope form.

7 0

2 years ago

A boat on a river travels downstream between two points, 90 mi apart, in 1 h. The return trip against the current takes 2 1 2 h.

docker41 [41]

Answer:

A)63miles per hour.

B)27 miles per hour

Step-by-step explanation:

HERE IS THE COMPLETE QUESTION

boat on a river travels downstream between two points, 90 mi apart, in 1 h. The return trip against the current takes 2 1 2 h. What is the boat's speed (in still water)??b) How fast does the current in the river flow?

Let the speed of boat in still water = V(boat)

speed of current=V(current)

To calculate speed of boat downstream, we add speed of boat in still water and speed of current. This can be expressed as

[V(boat) +V(current)]

It was stated that it takes 1hour for the

boat to travels between two points of 90 mi apart downstream.

To calculate speed of boat against current, we will substact speed of current from speed of boat in still water. This can be expressed as

[V(boat) - V(current)]

and it was stated that it takes 2 1/2 for return trip against the Current

But we know but Speed= distance/time

Then if we input the stated values we have

V(boat) + V(current)]= 90/1 ---------eqn(1)

V(boat) - V(current) = 90/2.5----------eqn(2)

Adding the equations we have

V(boat) + V(current) + [V(boat) - V(current)]= 90/2.5 + 90/1

V(boat) + V(current) + V(boat) - V(current)]=90+36

2V(boat)= 126

V(boat)=63miles per hour.

Hence, Therefore, the speed of boat in still water is 63 miles per hour.

?b) How fast does the current in the river flow?

the speed of the current in the river, we can be calculated if we input V(boat)=63miles per hour. Into eqn(1)

V(boat) + V(current)]= 90/1

63+V(current)=90

V(current)= 27 miles per hour

Hence,Therefore, the speed of current is 27 miles per hour.

7 0

2 years ago

A teacher claims that the proportion of students expected to pass an exam is greater than 80%. To test this claim, the teacher a

erica [24]

Answer:

z=-1.591

Step-by-step explanation:

Null Hypotheses,

$H_0 : p=0.8\\H_a: p>0.80$

So we use z-test for one population proportion (right-tailed test)

According to this informaton, we defined that

$\alpha=0.05$

$z_c=1.64$ (critical value)

So our Rejection region is $R={z:z>1.64}$

$z=\frac{\bar{p}-p_0}{\sqrt{p_0(1-p_0)/n}} = \frac{0.755-0.8}{\sqrt{0.8(1-0.8)/200}}=-1.591$

Not rejection.

6 0

2 years ago