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2 years ago

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In the diagram below, the length of the segment AB is 10, the length of segment AD is 20, and the length of segment AE is 18. Wh

at is the length of the segment AC?

1 answer:

DochEvi [55]2 years ago

4 0

The answer is d)9. Hope it helps

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I need help or the answer

loris [4]

Answer:

5x+4y = 52

Step-by-step explanation:

We can first write the equation in point slope form

y-y1 = m(x-x1) where m is the slope and (x1,y1) is the point

y - 8 = -5/4 ( x-4)

Multiply each side by 4 to get rid of the fraction

4(y - 8) = 4*(-5/4) ( x-4)

4(y - 8) = -5 ( x-4)

Distribute

4y - 32 = -5x+20

We want the equation in the form

Ax + By = C

Add 5x to each side

5x+4y -32 = -5x+5x+20

Add 32 to each side

5x+4y -32+32 =32+20

5x+4y = 52

4 0

2 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3><u>Required Answer</u><u>:</u><u>-</u></h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

2 years ago

What are the quotient and remainder of (3x^4+ 2x^2 - 6x + 1) /(x + 1)

quester [9]

Answer:

all work is shown and pictured

4 0

2 years ago

I need some help, an equation would do, will give you the Brainliest answer. Thank you!

kipiarov [429]

They will need 2 moving trucks

6 0

2 years ago

20 people applied for a job. Everyone either has a school certificate or diploma or even both. If 14 have school certificates an

STatiana [176]

Given:

Either has a school certificate or diploma or even both = 20 people

Having school certificates = 14

Having diplomas = 11

To find:

The number of people who have a school certificate only.

Solution:

Let A be the set of people who have school certificates and B be the set of people who have diplomas.

According to the given information, we have

$n(A)=14$

$n(B)=11$

$n(A\cup B)=20$

We know that,

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$20=14+11-n(A\cap B)$

$20=25-n(A\cap B)$

Subtract both sides by 25.

$20-25=-n(A\cap B)$

$-5=-n(A\cap B)$

$5=n(A\cap B)$

We need to find the number of people who have a school certificate only, i.e. $n(A\cap B')$ .

$n(A\cap B')=n(A)-n(A\cap B)$

$n(A\cap B')=14-5$

$n(A\cap B')=9$

Therefore, 9 people have a school certificate only.

3 0

2 years ago