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1 year ago

- How many terms are in -7x2 - 14x + 2 ?

Mathematics

2 answers:

Paladinen [302]1 year ago

7 0

Answer: 3

Explanation:
-7x2 first term
-14x second
2 third

Inessa05 [86]1 year ago

4 0

Answer:

Step-by-step explanation:

7x² is the first term,

14x is the second term and

2 is the third term.

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Bar chart question.. please help

vodomira [7]

D is the correct answer

3 0

3 years ago

In a collection of nickels, dimes, and quarters worth $6.90, the ratio of the number of nickels to dimes is 3:8. The ratio of th

dimulka [17.4K]

N=number of nickels
d=number of dimes
q=number of quarters

example
if n:d=3:8
lets say n=3 and d=8
3*8=8*3
so
8n=3d
slightly confusing but go with it, it migh make sense or it might not

n:d=3:8
8n=3d

d:q=4:5
5d=4q

solve for one of them
solve for nickels in first and quarters in second

8n=3d, n=(3/8)d
5d=4q, (4/5)d=q

5n+10d+25q=690
divide by 5 to make it easeir
n+2d+4q=138
sub (3/8)d for n
sub (4/5)d for q
(3/8)d+2d+5(5/4)d=138
(3/8)d+2d+25/4d=138

3/8+2+25/4=
3/8+16/2+50/8=
69/8

(69/8)d=138
times both sides by 8
69d=1104
divide both sides by 69
d=16

16 dimes
sub back

(5/4)d=q
(5/4)16=q
80/4=q
20=q

(3/8)d=n
(3/8)16=n
48/8=n
6=n

6 nickles
16 dimes
20 quarters
yah

8 0

3 years ago

An experiment observing the growth of two germ strand populations finds these patterns.

Mice21 [21]

Answer:

$R(x) = (1.3)^{3x-8}$

Step-by-step explanation:

Given

$A(x) = (1.3)^{x + 9}$

$B(x) = (1.3)^{4x + 1}$

Required

Ratio B(x) to A(x)

This is calculated as:

$R(x) = B(x) : A(x)$

Express as fraction

$R(x) = \frac{B(x) }{A(x)}$

Substitute: $A(x) = (1.3)^{x + 9}$ and $B(x) = (1.3)^{4x + 1}$

$R(x) = \frac{(1.3)^{4x+1}}{(1.3)^{x+9}}$

Apply law of indices

$R(x) = (1.3)^{4x-x+1-9}$

$R(x) = (1.3)^{3x-8}$

6 0

2 years ago

A binomial experiment has the given number of trials N and the given success probability P.

yan [13]

The probability of the success will be 0.3826.

<h3>What is binomial distribution?</h3>

Bernoulli's trials are those trials which end up randomly either on success (with probability p) or on failures (with probability 1- p = q (say))

The probability that out of n trials, there'd be x successes is given by

$\rm P(x) = \: ^nC_xp^x(1-p)^{n-x}$