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lara [203]
3 years ago
13

A number is 6 greater than 1/2 another number. If the sum of the numbers is 21, find the numbers

Mathematics
1 answer:
Anna71 [15]3 years ago
6 0

Answer:

x=11, y=10

Step-by-step explanation:

x= 6+ (1/2)y

x+y=21

substitute the first equation into the second equation:

(6+ (1/2y)) + y =21

to get:

y=10

then to find x:

x+ 10 = 21

x=11

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2y - 7+ 5y = 0<br> Step by step work
Ugo [173]

Answer:

y=1

Step-by-step explanation:

2y+5y-7=0

Combine y: 7y-7=0

Move 7 over: 7y=7

Divide by 7:y=1

5 0
3 years ago
Simplify 6-(-14)+(-2)​
garik1379 [7]

Answer:

18

Step-by-step explanation:

when there is a - in front of an expression in parentheses, change the sign of each term in the expression

6+14+(-2)

then calculate the sum or difference

6+14-2= 18

3 0
2 years ago
Giving brainlist to whoever answers
Setler79 [48]

Answer:

m<A=94

Step-by-step explanation:

(2x+10)+(2x+2)=180

4x+12=180

-12 -12

4x=168

/4 /4

x=42

m<A= 2x+10

m<A=2(42)+10

m<A=84+10

m<A=94

Check:

Add m<A and m<B which should equal 180.

m<B=2x+2

m<B=2(42)+2

m<B=84+2

m<B=86

94+86=180

3 0
2 years ago
A(1) = -11<br> a(n) = a(n − 1). 10<br> What is the 4th term in the sequence?
neonofarm [45]

Answer:

a(n)= a(n-1) .10

Clearly see the equation

you can do it directly Apply logic

Every n th term is 10 times of previous n-1 th term

That is common ratio an/a(n-1)= 10

So, We have to find 4 th term

So, 4 th term is simply

1st term × (common ratio)^(4-1)

-11( 10)^3

-11000

Thanks

3 0
3 years ago
Find an n^th degree polynomial with real coefficients satisfying the given conditions. n = 3; -2 and 2 i are zeros; f(-1) = 15.
Ira Lisetskai [31]
So, n = 3, is a 3rd degree polynomial, roots are -2 and 2i

well 2i is a complex root, or imaginary, and complex root never come all by their lonesome, their sister is always with them, the conjugate, so if 0+2i is there, 0-2i is there too

so, the roots are -2, 2i, -2i

now... \bf \begin{cases}&#10;x=-2\implies x+2=0\implies &(x+2)=0\\&#10;x=2i\implies x-2i=0\implies &(x-2i)=0\\&#10;x=-2i\implies x+2i=0\implies &(x+2i)=0&#10;\end{cases}&#10;\\\\\\&#10;(x+2)\underline{(x-2i)(x+2i)}=0\\\\&#10;-----------------------------\\\\&#10;\textit{difference of squares}&#10;\\ \quad \\&#10;(a-b)(a+b) = a^2-b^2\qquad \qquad &#10;a^2-b^2 = (a-b)(a+b)\\\\&#10;-----------------------------\\\\&#10;(x+2)[x^2-(2i)^2]=0\implies (x+2)[x^2-(2^2i^2)]=0&#10;\\\\\\&#10;(x+2)[x^2-(4\cdot -1)]=0\implies (x+2)(x^2+4)=0&#10;\\\\\\&#10;x^3+2x^2+4x+8=0

now, if we check f(-1), we end up with 5, not 15
hmmm

so, how to turn our 5 to 15? well, 3*5, thus

\bf 3(x^3+2x^2+4x+8)=f(x)\implies 3(5)=f(-1)\implies 15=f(-1)

usually, when we get the roots, or zeros, if any common factor that is a constant is about, they get in a division with 0 and get tossed, and aren't part of the roots, thus, we can simply add one, in this case, the common factor of 3, to make the 5 turn to 15
6 0
3 years ago
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