All categories

3 years ago

I need help with both of these. i need an answer fast please :)

Mathematics

1 answer:

r-ruslan [8.4K]3 years ago

4 0

6. 14/8=1.75
28/16=1.75
They are equal to each other

You might be interested in

Maybe can I have some help?

ivann1987 [24]

It’s -5 slope intercept is whatever point the line crosses the y axis

6 0

2 years ago

Choose the polynomial written in standard form.

Semenov [28]

Answer:

$x^4 + 4x^3 + 10x$

Step-by-step explanation:

A polynomial written in decreasing order of the degree of its monomials ( or single term ) is called its standard form,

In polynomial,

$x^2 + 4x^4 + 10x^6$ ,

Degrees are written in increasing order,

⇒ It is not written in standard form,

In polynomial,

$x^4 + 4x^3 + 10x$ ,

Degrees are written in decreasing order,

⇒ It is written in standard form,

In polynomial,

$x^7 + 4x^3 + 10x^4$ ,

There is no order of degrees,

⇒ It is not written in standard form,

In polynomial,

$x^6 + 4x^3 + 10x^7$ ,

There is no order of degrees,

⇒ It is not written in standard form,

4 0

2 years ago

Read 2 more answers

Solve 8x^2 = 65. Round to the nearest hundredth. the solutions are +/- _________

kirza4 [7]

2.850........,..,.hhh

7 0

2 years ago

A university hired you as a lab assistant after you graduated. Your starting salary is $41,500 and your salary will increase by

VladimirAG [237]

Answer:

$67,000

Step-by-step explanation:

41,000*(1,700*15)=67,000

5 0

3 years ago

Find (a) arc length and (b) Area of a sector.

barxatty [35]

Answer:

a) 29.45 cm (2 dp)

b) 220.89 cm² (2 dp)

Step-by-step explanation:

Formula

$\textsf{Arc length}=r \theta$

$\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta$

$\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}$

Calculation

Given:

$\theta=\dfrac{5 \pi}{8}$
r = 15 cm

$\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 15\left(\dfrac{5 \pi}{8}\right)\\& = \dfrac{75}{8} \pi \\& = 29.45\: \sf cm\:(2\:dp)\end{aligned}$

$\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(15^2) \left(\dfrac{5 \pi}{8}\right)\\\\& = \dfrac{225}{2}\left(\dfrac{5 \pi}{8}\right)\\\\ & = \dfrac{1125}{16} \pi \\\\& = 220.89 \: \sf cm^2\:(2\:dp)\end{aligned}$

5 0

1 year ago