All categories

3 years ago

Can someone help me plz

Mathematics

1 answer:

galina1969 [7]3 years ago

4 0

A= about 75 cents -- This is because you would figure out what 5% is equal to (.05). Then, you would do .05 x 14.90 to get the amount of sales tax you will be paying. By doing this you will get .745 which is approximately .75 which would become 75 cents

b= $15.65 -- For this one, you would do the cents (.75) plus the total (14.90) So you would do 14.90+.75 which = 15.65.

Hope this helped :)

You might be interested in

Please help! Question below

Marysya12 [62]

152... cuz 28 is perpendicular so... 180-28

5 0

4 years ago

Use the diagram and given information to answer the questions and prove the statement.

Greeley [361]

Answer:

See explanation

Step-by-step explanation:

ASA Postulate (Angle-Side-Angle):

If two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

Consider triangles XYB and ZYA. In these triangles

∠X≅∠Z (given)
XY≅ZY (given)
∠Y is common angle

By ASA Postulate, triangles XYB and ZYA are congruent. Congruent triangles have congruent corresponding sides, so

BX≅AZ

6 0

3 years ago

A scale model of a giant robot is 24 centimeters high with an arm length of 12 centimeters. The height of the real robot is 6 me

Travka [436]

The answer is 3. While scale real height is 24 arm is 12. 24/2=12
So6÷2=3

6 0

3 years ago

The graph below is the graph of a function:

arlik [135]

Answer:

Step-by-step explanation:

4 0

3 years ago

Use the definition of a Taylor series to find the first three non zero terms of the Taylor series for the given function centere

Ket [755]

Answer:

$e^{4x}=e^4+4e^4(x-1)+8e^4(x-1)^2+...$

$\displaystyle e^{4x}=\sum^{\infty}_{n=0} \dfrac{4^ne^4}{n!}(x-1)^n$

Step-by-step explanation:

Taylor series expansions of f(x) at the point x = a

$\text{f}(x)=\text{f}(a)+\text{f}\:'(a)(x-a)+\dfrac{\text{f}\:''(a)}{2!}(x-a)^2+\dfrac{\text{f}\:'''(a)}{3!}(x-a)^3+...+\dfrac{\text{f}\:^{(r)}(a)}{r!}(x-a)^r+...$

This expansion is valid only if $\text{f}\:^{(n)}(a)$ exists and is finite for all $n \in \mathbb{N}$ , and for values of x for which the infinite series converges.

$\textsf{Let }\text{f}(x)=e^{4x} \textsf{ and }a=1$

$\text{f}(x)=\text{f}(1)+\text{f}\:'(1)(x-1)+\dfrac{\text{f}\:''(1)}{2!}(x-1)^2+...$

$\boxed{\begin{minipage}{5.5 cm}\underline{Differentiating $e^{f(x)}$}\\\\If $y=e^{f(x)}$, then $\dfrac{\text{d}y}{\text{d}x}=f\:'(x)e^{f(x)}$\\\end{minipage}}$