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krek1111 [17]
3 years ago
9

Help i need help on this​

Mathematics
2 answers:
HACTEHA [7]3 years ago
8 0

Answer:

C. 27 cm^{2}

Step-by-step explanation:

For this problem find the area for the three individual segment then add them together.

<u>Step 1: Find area of the rectangle</u>

formula for area of a rectangle: bxh

b-base

h-height

3x6=18

<u>Step #2: Find area of the triangle</u>

Formula for area of a triangle:\frac{1}{2} bxh

b-base

h-height

\frac{1}{2} (3x3)=1/2(9)=4.5

<u>Step #3: Find the area of the triangles total</u>

Because the two triangles have the same area you can just multiply the area of one of the triangles by 2

4.5x2=9

<u>Step #4: Add all the areas together</u>

rectangle+2 triangles

18+9=27

liq [111]3 years ago
4 0

Answer:

27

Step-by-step explanation:

im almost possitive that that is right

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Trigonometry question
kaheart [24]

Answer:

c

Step-by-step explanation:

\tan \theta=\frac{perpendicular}{base} \\\tan 45=\frac{2200}{d} \\1=\frac{2200}{d} \\d=2200~ft.

8 0
3 years ago
Smith sold 45 of his DVDs and purchased 10 more. If Smith has 15 DVDs now, how many DVDs did he have to begin with?
Scorpion4ik [409]
Let's set up a formula for this, where x is the amount of dvds he started with:

x-45+10=15
x-35=15
 +35  +35
x=50

He had 50 DVDs to begin with.

Check
50-45+10
=15
6 0
3 years ago
Read 2 more answers
a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6?6x7x7=294 b) How many three-digit numbers
love history [14]

Answer:

a) 294

b) 180

c) 75

d) 168

e) 105

Step-by-step explanation:

Given the numbers 0, 1, 2, 3, 4, 5 and 6.

Part A)

How many 3 digit numbers can be formed ?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For unit's place, any of the numbers can be used i.e. 7 options.

For ten's place, any of the numbers can be used i.e. 7 options.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Total number of ways = 7 \times 7 \times 6 = <em>294 </em>

<em></em>

<em>Part B:</em>

How many 3 digit numbers can be formed if repetition not allowed?

Solution:

Here we have 3 spaces for the digits.

Unit's place, ten's place and hundred's place.

For hundred's place, 0 can not be used (because if 0 is used here, the number will become 2 digit) i.e. 6 options.

Now, one digit used, So For unit's place, any of the numbers can be used i.e. 6 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = 6 \times 6 \times 5 = <em>180</em>

<em></em>

<em>Part C)</em>

How many odd numbers if each digit used only once ?

Solution:

For a number to be odd, the last digit must be odd i.e. unit's place can have only one of the digits from 1, 3 and 5.

Number of options for unit's place = 3

Now, one digit used and 0 can not be at hundred's place So For hundred's place, any of the numbers can be used i.e. 5 options.

Now, 2 digits used, so For ten's place, any of the numbers can be used i.e. 5 options.

Total number of ways = 3 \times 5 \times 5 = <em>75</em>

<em></em>

<em>Part d)</em>

How many numbers greater than 330 ?

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 7

Number of options for unit's place = 7

Total number of ways = 3 \times 7 \times 7 = 147

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 7

Total number of ways = 1 \times 3 \times 7 = 21

Total number of required ways = 147 + 21 = <em>168</em>

<em></em>

<em>Part e)</em>

Case 1: 4, 5 or 6 at hundred's place

Number of options for hundred's place = 3

Number of options for ten's place = 6

Number of options for unit's place = 5

Total number of ways = 3 \times 6 \times 5 = 90

Case 2: 3 at hundred's place

Number of options for hundred's place = 1

Number of options for ten's place = 3 (4, 5, 6)

Number of options for unit's place = 5

Total number of ways = 1 \times 3 \times 5 = 15

Total number of required ways = 90 + 15 = <em>105</em>

7 0
3 years ago
If f(x) = x2 – 1 and g(x) = 2x – 3, what is the domain of (fºg)(x)?
RUDIKE [14]

Answer:

(-∞, ∞)

Step-by-step explanation:

Given,

f(x)=x^2-1 \ and\ g(x)=2x-3

Then (fºg)(x) would be ,

(2x-3)^2-1\\4x^2-12x+9-1\\4x^2-12x+8

Since this is a polynomial, we know domain of a polynomial is all real number.

Therefore the domain of this (fºg)(x) is (-∞, ∞).

5 0
3 years ago
I need help!! PLEASE PLEASEE HELP ME
zheka24 [161]
The second choice! I hope my math is right!
7 0
3 years ago
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