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2 years ago

PLEASE HELP !!!!!!!Find y, such that ABC ~A DEF.

Mathematics

1 answer:

Lemur [1.5K]2 years ago

7 0

Use proportionals to find y.

$\frac{9}{18} = \frac{6}{y}$

Answer:y=12

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PLEASE HELP!!!!!!!!!!!!!!!!

Anton [14]

Answer:

<h2>1. subtract 3x, subtract 4, divide by -4</h2><h2 /><h2>2. add x</h2>

Step-by-step explanation:

1 + 3x = -x + 4.

subtract 3x

1 = -4x + 4

subtract 4

-3 = -4x

divide by -4

(-3)/(-4) = (4x)/(-4) ⇌ x = 3/4

-x + 6 = 5 - 3x

subtract 5

-x + 1 = - 3x

add x

1 = -2x

4 0

3 years ago

Slope of a line perpendicular to 5x -9y = 1

Oksana_A [137]

I think it would be y=(-9/5)x -1/9

8 0

3 years ago

Julie is trying to choose between two packs of cookies. Pack A has 36 cookies for $5.90. Pack B has 24 cookies for $4.10. Which

algol13

Answer:

Step-by-step explanation:

7 0

3 years ago

Each treasure hunt team will have 5 people. So far 42 people have signed up. How many more people are needed to make every team

julia-pushkina [17]

Answer:

1) 3 more people

2) 9 teams

Step-by-step explanation:

The divisibility rule for 5 is that it must end in 5 and 0

so the next one is 45, so 3 more people

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

45/5=9 teams

3 0

1 year ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is:

$(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})$

$(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]$

$(x,y) = (10.452, 34.531)\,[km]$

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

$\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}$

$\theta \approx 16.840^{\circ}$

And the distance from the camp is calculated by the Pythagorean Theorem:

$r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}$

$r = 36.078\,km$

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0

3 years ago