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nikdorinn [45]
3 years ago
12

The similar figures, parallelograms ▱QUAD and ▱STOP, have a ratio of 3:1 between their corresponding sides. If = 15, then =

Mathematics
1 answer:
Artemon [7]3 years ago
4 0

Answer:

Ratio 3:1 or \frac{3}{1}. \frac{QUAD}{STOP}

AD = 15

OP = x

\frac{15}{x} = \frac{3}{1}

Cross multiply

15 × 1 = 3 · x

15 = 3x

\frac{15}{3} = x

5 = x

If AD = 15, then OP = 5

OP = 10

AD = x

\frac{x}{10} =\frac{3}{1}

Cross multiply

1 · x = 3 × 10

x = 30

If OP = 10, then AD = 30

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Llana [10]
<h2>Answer:</h2>

The vertex of the function is:

                             (-2,-2)

<h2>Step-by-step explanation:</h2>

We are given a absolute value function f(x) in terms of variable "x" as:

                       f(x)=-|x+2|-2

We know that for any absolute function of the general form:

f(x)=a|x-h|+k

the vertex of the function is : (h,k)

and if a<0 the graph of function opens downwards.

and if a>0 the graph of the function opens upwards.

Hence, here after comparing  the equation with general form of the equation we see that:

a= -1<0 , h= -2 and k= -2

Since a is negative , hence, the graph opens down .

         Hence, the vertex of the function is:

                         (-2,-2)

3 0
3 years ago
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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 <span>h'(x) 
</span>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

<span> h'(x) = 1.44 ------------ > not exist</span>

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fiasKO [112]

Answer:

see explanation

Step-by-step explanation:

Given

8x - 6 > 12 + 2x ( subtract 2x from both sides )

6x - 6 > 12 ( add 6 to both sides )

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Any value of x greater than 3 will be in the solution set

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How do you write 8.6 in scientific notation
aliina [53]

8.6

Since there is one number before the decimal, we don't have to move the decimal

8.6 * 10^0

Answer 8.6 * 10^0

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Which number is less than 3.58?​
sergiy2304 [10]

Answer:

3 is les than 3.58

Step-by-step explanation:

3 is less than 3.58

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