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3 years ago

RJ read 40 pages of gis book in 50 minutes. How many pages can I read in 80 minutes?

Mathematics

1 answer:

lisov135 [29]3 years ago

5 0

Answer:

64 pages

Step-by-step explanation:

40 pg = 50 min

8 pg = 10 min

64 pg = 80 min

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The student of astronauts u know that

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3 years ago

Please help!?<br><br> Evaluate the expression 5V - 6w when v= 4 and w= 2.

Rudik [331]

Answer:

Step-by-step explanation:

5V-6W

V=4. W=2.

5(4)-6(2)

20-6(2)

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3 years ago

Solve each equation 3m-3/5=2m-1/2

Kazeer [188]

Answer:

m= 1/10

Step-by-step explanation:

1. Move the terms so that will be 3m - 2m=-1/2 + 3/5

2. Subtract the m values so that will be 3m - 2m= 1 or m

3. Calculate the sum.

4. So the answer will be m= 1/10. M because we subtracted 2 from 3 which make us 1 so we say it m and 1/10 by doing 1/2 + 3/5.

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3 years ago

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docker41 [41]

Answer:

Step-by-step explanation:

V= BxHxL divided by 2

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3 years ago

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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