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3 years ago

The expression (3r+5)(2r-6) is equivalent to ?

Mathematics

1 answer:

Sholpan [36]3 years ago

5 0

Is the answer 5r-1? I'm not sure tho sorry if it's wrong!

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Tell whether the number 68 is a prime or composite number

stich3 [128]

68 is composite it has more factors than 1 and itself

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2 years ago

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Find an orderd pair (x,y) that is a solution to the equation 4x+y=9

Aleksandr [31]

Answer:

Below in bold.

Step-by-step explanation:

4x + y = 9

Let's put x = 0 then:

4(0) + y = 9

0 + y = 9

y = 9

So one. ordered pair that is a solution is (0, 9)

7 0

2 years ago

Show work and explain with formulas.

Juli2301 [7.4K]

20 Answer: a₁ = 4

Step-by-step explanation:

$a_n=324,\ r=3,\ n=5\\\\a_n=a_1 \cdot r^{n-1}\\\\324=a_1\cdot 3^{5-1}\\\\\dfrac{324}{3^4}=a_1\\\\\dfrac{324}{81}=a_1\\\\\large\boxed{4}=a_1$

21 Answer: n = 13

Step-by-step explanation:

$a_n=\dfrac{1}{64},\ a_1=64,\ r=\dfrac{1}{2}\\\\a_n=a_1 \cdot r^{n-1}\\\\\dfrac{1}{64}=64\cdot \bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{64\cdot 64}=\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\dfrac{1}{2^6\cdot 2^6}=\dfrac{1}{2^{n-1}}\\\\6+6=n-1\\\\\large\boxed{13}=n$

22 Answer: n = 5

Step-by-step explanation:

$a_n=48,\ a_1=1875\ r=\dfrac{2}{5}\\\\a_n=a_1 \cdot r^{n-1}\\\\48=1875\cdot \bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{48}{1875}=\bigg(\dfrac{2}{5}\bigg)^{n-1}\\\\\dfrac{16}{625}=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\\bigg(\dfrac{2}{5}\bigg)^4=\bigg(\dfrac{2}{5}\bigg)^{n-1}}\\\\4=n-1\\\\\large\boxed{5}=n$

6 0

3 years ago

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Please answer the question as the directions say and try your best :D

timurjin [86]

False: The relation is a function because for each X value there is only one Y value.

True: The domain is the x values
The lowest X value is 3 and the highest X value is 8
So 3<= x <= 8 is true

True: The range of a function is the Y values, which are 6,7,8,10

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2 years ago

Solve the equation for b. b + 34 - 34 = 130- does not equal 130.

aalyn [17]

Answer:

b+34-34=130

b+0=130

b=130-0

b=130

7 0

2 years ago