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kobusy [5.1K]
3 years ago
12

After reading 7/9 of a book, 40 pages are left. Find the total number of pages in the book​

Mathematics
1 answer:
dimaraw [331]3 years ago
6 0

Answer:

180 pages

Step-by-step explanation:

let x be the number of pages in the book , then

\frac{7}{9} x + 40 = x ( multiply through by 9 to clear the fraction )

7x + 360 = 9x ( subtract 7x from both sides )

360 = 2x ( divide both sides by 2 )

180 = x

The total number of pages in the book is 180

You might be interested in
In ΔBCD, the measure of ∠D=90°, the measure of ∠C=77°, and CD = 41 feet. Find the length of BC to the nearest foot.
OLga [1]

Answer:

182 ft

Step-by-step explanation:

When you are not given a diagram, always draw one for yourself (see diagram).

When you have a right triangle (triangle with 90° angle), you can use the trigonometry ratios, You can remember them using SohCahToa. It is read like this:

<u>s</u>inθ = <u>o</u>pposite/<u>h</u>ypotenuse           Soh

<u>c</u>osθ = <u>a</u>djacent/<u>h</u>ypotenuse          Cah

<u>t</u>anθ = <u>o</u>pposite/<u>a</u>djacent                Toa

"θ" means the angle of reference (the angle you are talking about).

We are looking for the length of BC, which is the <u>hypotenuse</u>. I labelled it "d" (lowercase D) because it is opposite to ∠D.

We know ∠C = 77°. This will be our angle of reference (replace θ).

The side we know is DC, also known as "b" (lowercase B) because it's opposite to ∠B. "b" is the <u>adjacent</u> side when θ = C because "b" is touching ∠C.

Take the general trig. formula that has <u>hypotenuse</u> and <u>adjacent</u>: (cosine ratio)

cosθ = adjacent/hypotenuse

Substitute the variables specific for this problem.

cosC = b/d

Substitute the values you know.

cos77° = (41 ft) / d

Isolate "d" to the left side

dcos77° = d*\frac{41ft}{d}     Multiply both sides by "d"

dcos77° = 41 ft                          

dcos77° / cos77° = 41 ft / cos77°        Divide both sides by cos77°

d = 41 ft / cos77°                      Input into calculator

d = 182.261874....... ft           Unrounded answer

d ≈ 182 ft                       Rounded to nearest foot (whole number)

Remember d = BC. It's often easier to use one letter for calculations.

Therefore the length of BC is about 182 feet.

8 0
3 years ago
Which of the following describes a net of the prism
Grace [21]

The answer is C

Explanation; there are 2 triangles and 3 rectangles

7 0
3 years ago
Read 2 more answers
What are some real life situations using system of equations?
OlgaM077 [116]
Bob's car rental company makes you pay 10 dollars per day you rent the car and a 30 dollar insurance fee

joe's car rental company makes you pay 30 dollars per day you rent the car and a 10 dollar insurance fee

how many days do you need to rent a car for the cost for renting both are the same

bob=10x+30
joe=30x+10
set each to each other
10x+30=30x+10
subtract 10x
30=20x+10
subtract 10
20=20x
divide both sides by 20
1=x
you need to rent 1 day for them to be equal
3 0
3 years ago
Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.
alina1380 [7]

Answer:

v_1=(\frac{1}{10},-\frac{3}{10})

v_2=(-\frac{1}{10},\frac{3}{10})

Step-by-step explanation:

First we define two generic vectors in our \mathbb{R}^2 space:

  1. v_1 = (x_1,y_1)
  2. v_2 = (x_2,y_2)

By definition we know that Euclidean norm on an 2-dimensional Euclidean space \mathbb{R}^2 is:

\left \| v \right \|= \sqrt{x^2+y^2}

Also we know that the inner product in \mathbb{R}^2 space is defined as:

v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

\left \| v_1 \right \|= \sqrt{x^2+y^2}=1

and

\left \| v_2 \right \|= \sqrt{x^2+y^2}=1

As second condition we have that:

v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0

v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0

Which is the same:

y_1=-3x_1\\y_2=-3x_2

Replacing the second condition on the first condition we have:

\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}

Since x_1^2= \frac{1}{10} we have two posible solutions, x_1=\frac{1}{10} or x_1=-\frac{1}{10}. If we choose x_1=\frac{1}{10}, we can choose next the other solution for x_2.

Remembering,

y_1=-3x_1\\y_2=-3x_2

The two vectors we are looking for are:

v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})

5 0
3 years ago
What can you say about the two figures
Tomtit [17]
The answer would be letter D!

hope this helps!! pls also mark me brainliest if you can :)) i am trying to level up. thank you!!
6 0
3 years ago
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