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iogann1982 [59]
3 years ago
13

The cost (in dollars) in making b bracelets is represented by 4+5b. The cost (in dollars) of making b necklaces is represented b

y 8b+6. Write a polynomial in standard form that represents how much more it costs to make b necklaces than b bracelets.
Mathematics
1 answer:
ddd [48]3 years ago
8 0
Answer:

P(b)=3b+2

Step-by-step explanation:

The cost (in dollars) in making b bracelets is represented is

B(b)=4+5b

The cost (in dollars) of making b necklaces is

N(b)=8b+6

To find how much more it will cost to make b necklaces than b bracelets, we find the difference,

P(b)=N(b)-B(b)

P(b)=8b+6-(4+5b)

P(b)=8b+6-4-5b

P(b)=3b+2

This is a polynomial in descending powers of b, hence it is in standard form;
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Can someone help me with these problems please! Im so confused
Schach [20]

Step-by-step explanation:

why are you confused ?

the question is to get the area of the shaded (= grey) areas.

all the shapes are clearly combined shapes out of several basic shapes, and in some cases we need to add then, and in some we need to subtract them from each other (to eliminate any white content).

that is really all that needs to be done.

of course, it is important to notice the different metrics used for the lengths, which define the metrics to be used for the areas.

a.

a rectangle and 2 half-circles on the sides (they are together one full circle).

radius is always 1/2 of the diameter of a circle.

and the width of the rectangle is exactly the diameter (= 2×radius) of the circle(s).

grey area :

rectangle : 20×(6×2) = 20×12 = 240 m²

circle : pi×r² = pi×6² = pi×36 = 113.0973355... m²

in total :

240 + 113.0973355... = 353.0973355... m²

b.

a trapezoid minus a circle.

the area of a trapezoid is

(a+b)×h/2

and h (height) is again the diameter of the circle (2×radius).

grey area :

trapezoid : (9 + 15)(4×2)/2 = 24 × 4 = 96 ft²

circle : pi×r² = pi×4² = pi×16 = 50.26548246... ft²

in total :

96 - 50.26548246... = 46.26548246... ft²

c.

3 rectangles.

rectangle1 - rectangle2 + rectangle3.

grey area :

rectangle1 : 14×10 = 140 m²

rectangle2 : 10×6 = 60 m²

rectangle3 : 6×2 = 12 m²

in total :

140 - 60 + 12 = 92 m²

d.

a circle minus a square.

the only specialty here is the diameter and radius of the circle : the diagonal of the inner square.

we get the diagonal of the inner square via Pythagoras, because the diagonal makes 2 right-angled triangles out of the square.

diagonal² = 4² + 4² = 16+16 = 32

diagonal = sqrt(32)

radius = diagonal/2 = sqrt(32)/2 = sqrt(32/4) = sqrt(8)

grey area :

square : 4×4 = 16 cm²

circle : pi×r² = pi×(sqrt(8))² = pi×8 = 25.13274123... cm²

in total :

25.13274123... - 16 = 9.132741229... cm²

is there still anything unclear ?

please let me know.

but as you can see there is no need to panic. just use basic math and common sense.

8 0
2 years ago
Think of one leg of a right triangle is 21' and the length of the hypothenuse is 29' what is the length of the other leg
Ksivusya [100]

Answer:

length of other leg = 20'

Step-by-step explanation:

Given:

One leg of triangle let it be base of triangle = 21'

Hypotenuse = 29'

To Find:

Third leg =perpendicular = ?

Solution:

Given is a right angled triangle with two of its sides given to us

We have Pythagoras theorem for right angled triangle and according to it

Pythagoras theorem is

for a right angled triangle

(hyp) ² = (perp)²+(base)²

We have the values of hyp and base

putting them in the equation we get

(29) ² = (perp) ²+(21) ²

841 = (perp) ² + 441

subtracting 441 from both sides we get

841 - 441 = (perp) ² + 441 - 441

400 =  (perp)²

or

(perp) ²=400

Taking square root of both sides

\sqrt{(perp)^{2} }=\sqrt{400}

it gives us

perp = ± 20'

As it is length and it can not be negative so

perp = 20'

or

length of other leg = 20'

3 0
3 years ago
Which statement is true regarding the functions on the graph?
erastova [34]

Answer:

The answer to the questions Is f3=g6

8 0
3 years ago
How do compare ratios then simplify to find the answer
gogolik [260]

"Compare ratios" covers a lot of territory. If all you want to do is find which one is larger, you can subtract or divide.

(a/b) - (c/d) > 0 . . . . means a/b is larger

(a/b) / (c/d) > 1 . . . . means a/b is larger

These operations with fractions are done using the methods of arithmetic with fractions that you have been taught.

____

"Simplify" when applied to ratios usually means common factors are removed from each of the terms.

For example, the ratio 2:12 is a ratio of even numbers, so we know both numbers have a factor of 2. (2 is a factor common to both numbers.) When we divide them both by 2, we have the reduced ratio 1:6. That is, 2:12 simplifies to become 1:6.

(It is helpful to have a good working knowledge of multiplication tables when you approach problems in simplifying ratios.)

= = = = = = = = = =

In "quick and dirty" terms, you can do the subtraction ...

\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-bc}{bd}

That is, you really only need to find out if (ad) > (bc) to determine if (a/b) > (c/d). A similar result is obtained when you consider division ...

\displaystyle\frac{\left(\frac{a}{b}\right)}{\left(\frac{c}{d}\right)}=\frac{a}{b}\cdot\frac{d}{c}=\frac{ad}{bc}

This will be greater than 1 if (ad) > (bc), signifying that (a/b) > (c/d).

_____

Here's an example with numbers.

... Compare 6/7 to 43/50.

... The relationship of interest will be revealed by the products 6·50 = 300 and 7·43 = 301.

... Since 300 < 301, these tell us 6/7 < 43/50.

_____

We say the above methods are "quick and dirty" because the results may need to be simplified when you're done.

For example, subtracting 1/6 from 1/2 gives

... 1/2 - 1/6 = (1·6 - 2·1)/(2·6) = 4/12

Here, numerator and denominator have a common factor of 4. Removing that gives 4/12 = 1/3.

Even if you do this by the method of common denominators, you still need to reduce the result.

... 1/2 - 1/6 = 3/6 - 1/6 = (3-1)/6 = 2/6

This must be reduced to 1/3 by removing a common factor of 2 from numerator and denominator.

3 0
3 years ago
Can you help me with my school work
MissTica

Answer:

I guess. Just post question.


8 0
3 years ago
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