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lidiya [134]
3 years ago
5

Someone please help me, I need the type of statement and reason.

Mathematics
1 answer:
RideAnS [48]3 years ago
5 0
The type of statement is a congruent
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PLS HELP THIS IS HARD ANYONE PLS
aalyn [17]

Answer:

a or c

Step-by-step explanation:

4 0
3 years ago
The mean of 10 values is 19. If further 5 values are
Leokris [45]

Answer:

22

Step-by-step explanation:

Pretend the 10 values in the first sentence are a,b,c,d,e,f,g,h,i,j

Pretend the addition 5 values is k,l,m,n,o

So the mean of all the 15 data is (a+b+c+d+e+f+g+h+i+j+k+l+m+n+o)/15=20

So the sum of all 15 data is  a+b+c+d+e+f+g+h+i+j+k+l+m+n+o=300                since 15(20)=300

Now let's look at the first 10:  We have their mean so we can write:

(a+b+c+d+e+f+g+h+i+j)/10=19

so a+b+c+d+e+f+g+h+i+j=190                               since 10(19)=190

So that means using our first sum equation and our equation sum equation we have

190+k+l+m+n+o=300

      k+l+m+n+o=300-190

      k+l+m+n+o= 110

So the average of those 5 numbers mentioned in your problem is 110/5=22

3 0
3 years ago
What is 4/6 equal to
timama [110]

\frac{2}{3}
8 0
3 years ago
Verify the identity. 4 csc 2x = 2 csc2x tan x
vlada-n [284]

Step-by-step explanation:

4csc(2x) = 2csc^2(x) tan(x)

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

4 \frac{1}{sin(2x)}

Also we use identity

sin(2x) = 2 sin(x) cos(x)

4 \frac{1}{2sin(x)cos(x)}

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}

2\frac{sin(x)}{sin^2(x)cos(x)}

2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}

We know that sinx/ cosx = tan(x)

Also  1/ sin(x)= csc(x)

so it becomes 2csc^2(x) tan(x) , Right hand side

Hence verified



6 0
3 years ago
Multiply.<br> (d+7)(d-7)<br> Simplify your answer.
romanna [79]

Answer:

\huge\boxed{(d+7)(d-7)=d^2-49}

Step-by-step explanation:

\bold{METHOD\ 1:}\\\\(d+7)(d-7)\qquad|\text{use}\ (a+b)(a-b)=a^2-b^2\\\\=d^2-7^2=d^2-49\\\\\bold{METHOD\ 2:}\\\\(d+7)(d-7)\qquad|\text{use FOIL}\ (a+b)(c+d)=ac+ad+bc+bd\\\\=(d)(d)+(d)(-7)+(7)(d)+(7)(-7)\\\\=d^2-7d+7d-49\qquad|\text{combine like terms}\\\\=d^2+(-7d+7d)-49\\\\=d^2-49

7 0
3 years ago
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