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hoa [83]
3 years ago
7

What is the value of x in a & b?

Mathematics
1 answer:
lina2011 [118]3 years ago
3 0

Answer:

a. x = 9

b. x = 3.6

Step-by-step explanation:

a. By Basic proportionality theorem:

AB/BD = AC/CE

4/4 = x/9

1 = x/9

9 = x

x = 9

b.

In \: \triangle ABC \: \& \: \triangle EDF\\\angle B \cong \angle D....(given) \\\angle C \cong \angle F.....(given) \\\therefore \triangle ABC \: \sim \: \triangle EDF.. (AA\: POSTULATE) \\\therefore \frac{AB}{ED} = \frac{BC}{DF} ...(csst) \\\therefore \frac{2}{5} = \frac{x}{9}\\\therefore x = \frac{2\times 9}{5} \\\therefore x = \frac{18}{5} \\\huge \red{\boxed {\therefore x = 3.6}} \\

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Find four consecutive even integers whose sum is 84. Find the integers.
mina [271]
Suppose the integers are n , n+2 , n+4 and n+6.
84=n+(n+2)+(n+4)+(n+6)=4n+12.
Subtract 12 from both ends to get.
72=4n.
Divide both ends by 4 to get.
n=18.
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4 0
3 years ago
Ted knows the actual distance between two cities is 150 miles. His map shows a distance of 2 inches between these cities. What i
Serjik [45]

Answer:

The scale used on his map is <u>150 miles : 2 inches</u>.

Step-by-step explanation:

Given:

Ted knows the actual distance between two cities is 150 miles. His map shows a distance of 2 inches between these cities.

Now, to find the scale Ted used on his map.

The actual distance Ted know between two cities = 150 miles.

The distance on map between these cities = 2 inches.

So, to get the scale used on his map:

\frac{The\ actual\ distance\ Ted\ know\ between\ two\ cities}{The\ distance\ on\ map\ between\ these\ cities}

=\frac{150\ miles}{2\ inches}

=150\ miles:2\ inches.

Therefore, the scale used on his map is 150 miles : 2 inches.

5 0
2 years ago
Select all the pieces of the triangle that are marked as congruent in the following picture
nikitadnepr [17]

Answer:

C. D. E. is the correct answer.

Step-by-step explanation:

Hope this helps you. Have a nice day. If you have doubt then please ask in comments.

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So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.

2. Asymptotes:

f(x)=\dfrac{x-3}{x^2-2x-3}=\dfrac{x-3}{(x-3)(x+1)}=\boxed{\dfrac{1}{x+1}}

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Check the picture below.

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