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3 years ago

Explain why the difference of two polynomials will always be a polynomial

1 answer:

8 0

<span>They are finite sums of expressions of the form where a is a constant, x is the variable and k is a non-negative integer</span>

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I don’t understand the question and I can’t find anything in my notes to help

lara [203]

Step-by-step explanation:

The problem states that you have a linear function so expect your equation to have this form:

y = mx + b

where m is the slope and b is the y-intercept. You are also given two points: P1(5, 6) and P2(14, 60). Use these points to solve for the slope m.

m = (y2 - y1) / (x2 - x1) = (60 - 6)/(14 - 5)

= 54/9 = 6

So our equation now becomes

y = 6m + b

To solve for b, plug in the values of P1:

6 = 6(5) + b ---> b = -24

Therefore, our equation is

y = 6m - 24

The rest of the points are

(8, 24)

(11, 42)

7 0

3 years ago

Help with my homework. Please.

garik1379 [7]

Answer:

Step-by-step explanation: since average speed = total distance/time

a)average speed of the journey from my home town to A = 10 miles/20 minutes = 0.5

b)journey from my home town to B = 20 miles/30 minutes = 0.7

c)10 miles + 20 miles = 30 miles

which takes 50 minutes

so: 30 miles/50 minutes = 0.6

5 0

3 years ago

When taking skinfold measurement readings only one attempt per site is necessary for an accurate reading.a. Trueb. False

Eva8 [605]

Answer: False

Step-by-step explanation: Skinfold measurements is one of the oldest ways of measuring a person's fat percentage,it's usually taken in specific areas of the body where there are Skinfolds,while taking this measurements it is expected that the person taking it does the average of 2or more repeated measurements in order to ensure that the actual thickness of that area of the body is correctly entered.

It is specifically taken from the right side of the body,where the person pinches out the Skinfolds away from the body by attaching a caliper ,this is to ensure that only the fatty laters are considered, it is mainly presented in percentage.

3 0

3 years ago

The dollar value

Hoochie [10]

Hi there
The formula you meant
V (t)=27500 (0.84)^t
V (10)=27,500×(0.84)^(10)
V (10)=4,809.8 Round your answer to get 4810 ...this isthe value after
10 years

the initial value of the car is
27500

Hope it helps

4 0

3 years ago

1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.

m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = $\frac{9}{7+9}\times (x)$

= $\frac{9}{16}\times (x)$

And the weight of zinc = $\frac{7}{7+9}\times (x)$

= $\frac{7}{16}\times (x)$

If the weight of zinc = 31.5 kg

31.5 = $\frac{7}{16}\times (x)$

x = $\frac{16\times 31.5}{7}$

x = 72 kgs

Therefore, weight of copper = $\frac{9}{16}\times (72)$

= 40.5 kgs

2). i). 2 : 3 = $\frac{2}{3}$

4 : 5 = $\frac{4}{5}$

Now we will equalize the denominators of each fraction to compare the ratios.

$\frac{2}{3}\times \frac{5}{5}$ = $\frac{10}{15}$

$\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}$

Since, $\frac{12}{15}>\frac{10}{15}$

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = $\frac{11}{19}$

19 : 21 = $\frac{19}{21}$

By equalizing denominators of the given fractions,

$\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}$

And $\frac{19}{21}\times \frac{19}{19}=\frac{361}{399}$

Since, $\frac{361}{399}>\frac{231}{399}$

Therefore, 19 : 21 > 11 : 19

iii). $\frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}$

$=\frac{3}{2}$

$\frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}$

= $\frac{4}{3}$

Now we equalize the denominators of the fractions,

$\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$

And $\frac{4}{3}\times \frac{2}{2}=\frac{8}{6}$

Since $\frac{9}{6}>\frac{8}{6}$

Therefore, $\frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4}$ will be the answer.

IV). $1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}$

$=\frac{6}{5}\times \frac{3}{4}$

$=\frac{18}{20}$

$=\frac{9}{10}$

Similarly, $\frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}$

$=\frac{4}{15}$

By equalizing the denominators,

$\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}$

Similarly, $\frac{4}{15}\times \frac{20}{20}=\frac{80}{300}$

Since $\frac{270}{300}>\frac{80}{300}$

Therefore, $1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}$

V). If a : b = 6 : 5

$\frac{a}{b}=\frac{6}{5}$

$=\frac{6}{5}\times \frac{2}{2}$

$=\frac{12}{10}$

And b : c = 10 : 9

$\frac{b}{c}=\frac{10}{9}$

Since a : b = 12 : 10

And b : c = 10 : 9

Since b = 10 is common in both the ratios,

Therefore, combined form of the ratios will be,

a : b : c = 12 : 10 : 9

7 0

3 years ago