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3 years ago

Find the area of the SHADED region.

Mathematics

1 answer:

7nadin3 [17]3 years ago

6 0

Split the shape into two shapes

A square

Area = L x H = 20 x 21 = 240 cm2

A right angle triangle

Area of right angle triangle = L x H divided by 2

20 x 21 = 240 divided by 2 = 120 cm2

Add both areas

240 + 120 = 360 cm2

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What is the slope of the line in the graph?

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Factor using the trinomial 10. X2 + 13x + 12

Rudik [331]

Answer:

Factoring the trinomial: $x^2+13x+12$ we get the factors $\mathbf{(x+1)(x+12)}$

Step-by-step explanation:

We need to factor the trinomial: $x^2+13x+12$

We can factor the trinomial by breaking the middle term i.e 13x

Such that adding or subtracting gets the both terms we get 13x and multiplying both terms we get 12x^2

We can break 13x as: 12x and x

Adding them we get 13x and multiplying them we get 12x^2

So,

$x^2+13x+12\\=x^2+12x+x+12\\=x(x+12)+1(x+12)\\=(x+1)(x+12)$

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8 0

2 years ago

5. What is the center of (0.1, 0.3, 0.02, 0.09, 0.4, 0.05, 0.26}? 0.2 0.3 0.1 0.4

Dima020 [189]

Answer:

0.09

Step-by-step explanation:

The data set is:

0.1, 0.3, 0.02, 0.09, 0.4, 0.05, 0.26

The number in the middle is 0.09. Hope this helps!

3 0

3 years ago

Can I get help solving this graph please?

Daniel [21]

see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find h'(x)
h'(x)=-36/25=-1.44

6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

 h'(x) = 1.44 ------------ > not exist

8 0

2 years ago