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3 years ago

8

Someone please help me

1 answer:

dem82 [27]3 years ago

6 0

Which problem? I will help you if you tell me what problem it is.

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write the equation of the line parallell to the given line passing through the given point in slope intercept form y=-4x-2 (0, -

Finger [1]

Step-by-step explanation:

slope: -4

y + 1 = -4(x - 0)

y + 1 = -4x + 0

y = -4x - 1

4 0

3 years ago

Solve the proportion 20/3=?/6

Zepler [3.9K]

It equals 40/6. I hope this helps. Good luck!

7 0

3 years ago

Read 2 more answers

Find the measure of each missing angle.

nadya68 [22]

Answer:

Step-by-step explanation:

Angle sum property of triangle: Sum of all the angles of a triangle is 180

Alternate interior angles: When two parallel lines are intersected by a transversal, the pair of angles on the inner side of each of these lines but on the opposite side of the transversal are called alternate interior angles

In ΔABC,

∠1 + 90 + 38 = 180 {angle sum property of triangle}

∠1 + 128 = 180

∠1 = 180 - 128

$\sf \boxed{\bf \angle 1 = 52^ \°}$

AB // CD and AC is transversal.

$\sf \boxed{\bf \angle 3 = 38^\circ}$ {Alternate interior angles are equal}

In ΔACD,

∠2 + ∠3 + 63 = 180 {angle sum property of triangle}

∠2 + 38 +63 = 180

∠2 + 101 =180

∠2 = 180 - 101

$\sf \boxed{\bf \angle 2 = 79^ \°}$

5 0

2 years ago

What is the length of BC , rounded to the nearest tenth?

Arte-miy333 [17]

Step $1$

In the right triangle ADB

<u>Find the length of the segment AB</u>

Applying the Pythagorean Theorem

$AB^{2} =AD^{2}+BD^{2}$

we have

$AD=5\ units\\BD=12\ units$

substitute the values

$AB^{2}=5^{2}+12^{2}$

$AB^{2}=169$

$AB=13\ units$

Step $2$

In the right triangle ADB

<u>Find the cosine of the angle BAD</u>

we know that

$cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}$

Step $3$

In the right triangle ABC

<u>Find the length of the segment AC</u>

we know that

$cos(BAC)=cos (BAD)=\frac{5}{13}$

$cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{AB}{AC}$

$\frac{5}{13}=\frac{13}{AC}$

solve for AC

$AC=(13*13)/5=33.8\ units$

Step $4$

<u>Find the length of the segment DC</u>

we know that

$DC=AC-AD$

we have

$AC=33.8\ units$

$AD=5\ units$

substitute the values

$DC=33.8\ units-5\ units$

$DC=28.8\ units$

Step $5$

<u>Find the length of the segment BC</u>

In the right triangle BDC

Applying the Pythagorean Theorem

$BC^{2} =BD^{2}+DC^{2}$

we have

$BD=12\ units\\DC=28.8\ units$

substitute the values

$BC^{2}=12^{2}+28.8^{2}$

$BC^{2}=973.44$

$BC=31.2\ units$

therefore

<u>the answer is</u>

$BC=31.2\ units$

8 0

3 years ago

Read 2 more answers

Please respond to the question below in your own words with proper explanations

ycow [4]

Answer:

$t(400)=673$ .

It means that 673 of 9 inch tiles are required to cover an area of 400 square feet, when spacing the tiles a quarter inch apart.

Step-by-step explanation:

Given:

The function relating number of 9 inch tiles required and area to cover is:

$t(A)=\frac{144}{9.25^2}A$

Now, plug in 400 for $A$ to evaluate $t(400)$ .

$t(400)=\frac{144}{9.25^2}\times 400=\frac{144\times 400}{85.5625}=\frac{57600}{85.5625}=673.19\approx6225$

Therefore, Don required 673 of 9 inch tiles to cover an area of 400 square feet, when spacing the tiles a quarter inch apart.

4 0

3 years ago