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2 years ago

5

HELP PLEASE 40 POINTS

1 answer:

Natali [406]2 years ago

5 0

Answer:

the answer is A

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Nathaniel participated in an 18-mile race. He ran 1 2 mile in 1 6 hour. If he ran the entire race at the same pace, how long, in

Softa [21]

Answer:

24 hours

Step-by-step explanation:

set up a ratio:

12 miles = 16 hours

------------- ------------

18 miles = x hours

cross multiply

12m * x = 16h * 18 m

solve for x

x=16h * 18m / 12m = 24 hours

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3 years ago

A coin which is tossed straight up into the air. After it is released it moves upward, reaches its highest point and falls back

Helen [10]

Answer:

Answers G, and A

Step-by-step explanation:

the motion of a body is as a result of the various forces acting on it.

As the coin is tossed upwards an upwards force acts on it and this force decreases up to the highest point the coin reaches. at the highest point the force of gravity applies. This force is downwards and constant (9.8m/ $s^{2}$ ).

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3 years ago

Select all the expressions that are equivalent to (2)^n+³

eimsori [14]

Answer:

The expressions which equivalent to $(2)^{n+3}$ are:

$4(2)^{n+1}$ ⇒ B

$8(2)^{n}$ ⇒ C

Step-by-step explanation:

Let us revise some rules of exponent

$a^{m}$ × $a^{m}$ = $a^{m+n}$
$(a^{m})^{n}$ = $a^{m*n}$

Now let us find the equivalent expressions of $(2)^{n+3}$

A.

∵ 4 = 2 × 2

∴ 4 = $2^{2}$

∴ $(4)^{n+2}$ = $(2^{2})^{n+2}$

- By using the second rule above multiply 2 and (n + 2)

∵ 2(n + 2) = 2n + 4

∴ $(4)^{n+2}$ = $(2)^{2n+4}$

B.

∵ 4 = 2 × 2

∴ 4 = 2²

∴ $4(2)^{n+1}$ = 2² × $(2)^{n+1}$

- By using the first rule rule add the exponents of 2

∵ 2 + n + 1 = n + 3

∴ $4(2)^{n+1}$ = $(2)^{n+3}$

C.

∵ 8 = 2 × 2 × 2

∴ 8 = 2³

∴ $8(2)^{n}$ = 2³ × $(2)^{n}$

- By using the first rule rule add the exponents of 2

∵ 3 + n = n + 3

∴ $8(2)^{n}$ = $(2)^{n+3}$

D.

∵ 16 = 2 × 2 × 2 × 2

∴ 16 = $2^{4}$

∴ $16(2)^{n}$ = $2^{4}$ × $(2)^{n}$

- By using the first rule rule add the exponents of 2

∵ 4 + n = n + 4

∴ $16(2)^{n}$ = $(2)^{n+4}$

E.

$(2)^{2n+3}$ is in its simplest form

The expressions which equivalent to $(2)^{n+3}$ are:

$4(2)^{n+1}$ ⇒ B

$8(2)^{n}$ ⇒ C

3 0

3 years ago

If a Customer places an order for $47.36 and it’s late, then how much would the customer be refunded if the refund is 20% on the

Serggg [28]

$9.47

I just multiplied the total by the refund percentage to get that number. It’s just like finding a tip.

6 0

2 years ago

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago