All categories

3 years ago

Answer the photo below thanks

Mathematics

2 answers:

nataly862011 [7]3 years ago

7 0

Answer:

octagon, because 4 + 4 = eight sides, and octagons have 8 sides

Step-by-step explanation:

Hope that helps!

beks73 [17]3 years ago

4 0

Answer:

B) octagon

Step-by-step explanation:

4 corners +4 corners =8 corners (or 8 sides)

You might be interested in

A student purchases school supplies at a store where the sale tax is 7%. The pre-tax price of the supplies is $48. What is the t

Alenkasestr [34]

Solution: We are given:

Sales tax $=7\%$

The pre-tax price of the supplies $= 48$

We have to find the total cost of the supplies.

We first need to find the sales tax on the price of supplies. The sales tax amount is:

$7\% of 48=0.07 \times 48=3.36$

Therefore, the total cost of the supplies = pre-tax price of the sales + sales tax amount

=$48 + $3.36

=$51.36

6 0

3 years ago

Which frame processing method causes a switch to wait until the first 64 bytes of the frame have been received before forwarding

malfutka [58]

Fragment free frame processing method causes a switch to wait until the first 64 bytes of the frame have been received before forwarding the frame to the destination device

Fragment free processing

An advanced type of cut-through switching is fragment-free (runtless switching). Before making a switching choice, switches engaged in cut-through switching merely read the Ethernet frame's destination MAC address field.

What distinguishes fragment-free switching from fast forward switching?

The latency in fast-forward mode is calculated between the first bit received and the first bit delivered. The most common cut-through switching technique is fast-forward switching. Switching without fragments: The switch stores the first 64 bytes of the frame in fragment-free switching before forwarding.

Learn more about frame here: brainly.com/question/19756844

#SPJ4

3 0

2 years ago

Please help! Thx (ASAP)

OverLord2011 [107]

Answer:

B. 3 to 10

Step-by-step explanation:

She uses 3 scoops of hot cocoa.

She uses 10 miniature marshmallows.

The ratio of scoops of hot cocoa to miniature marshmallows is 3 to 10.

6 0

3 years ago

Let R be the region bounded by

loris [4]

a. The area of $R$ is given by the integral

$\displaystyle \int_1^2 (x + 6) - 7\sin\left(\dfrac{\pi x}2\right) \, dx + \int_2^{22/7} (x+6) - 7(x-2)^2 \, dx \approx 9.36$

b. Use the shell method. Revolving $R$ about the $x$ -axis generates shells with height $h=x+6-7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le 2$ , and $h=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With radius $r=x$ , each shell of thickness $\Delta x$ contributes a volume of $2\pi r h \Delta x$ , so that as the number of shells gets larger and their thickness gets smaller, the total sum of their volumes converges to the definite integral

$\displaystyle 2\pi \int_1^2 x \left((x + 6) - 7\sin\left(\dfrac{\pi x}2\right)\right) \, dx + 2\pi \int_2^{22/7} x\left((x+6) - 7(x-2)^2\right) \, dx \approx 129.56$

c. Use the washer method. Revolving $R$ about the $y$ -axis generates washers with outer radius $r_{\rm out} = x+6$ , and inner radius $r_{\rm in}=7\sin\left(\frac{\pi x}2\right)$ if $1\le x\le2$ or $r_{\rm in} = 7(x-2)^2$ if $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each washer has volume $\pi (r_{\rm out}^2 - r_{\rm in}^2) \Delta x$ . As more and thinner washers get involved, the total volume converges to

$\displaystyle \pi \int_1^2 (x+6)^2 - \left(7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \pi \int_2^{22/7} (x+6)^2 - \left(7(x-2)^2\right)^2 \, dx \approx 304.16$ <em />

d. The side length of each square cross section is $s=x+6 - 7\sin\left(\frac{\pi x}2\right)$ when $1\le x\le2$ , and $s=x+6-7(x-2)^2$ when $2\le x\le\frac{22}7$ . With thickness $\Delta x$ , each cross section contributes a volume of $s^2 \Delta x$ . More and thinner sections lead to a total volume of

$\displaystyle \int_1^2 \left(x+6-7\sin\left(\frac{\pi x}2\right)\right)^2 \, dx + \int_2^{22/7} \left(x+6-7(x-2)^2\right) ^2\, dx \approx 56.70$

7 0

2 years ago

Examination, 152 Candidates offered Agriculture

agasfer [191]

SOLUTION

Given the question in the image, the following are the solution steps to answer the question

STEP 1: Write the given details

$n(G)=136,n(P)=128,n(A)=152,n(A\cap G)=52,n(A\cap P)=60,n(A\cap P\cap G)=30,n(P\cap G)=58$

STEP 2: Get the number of students that takes the subjects only

From the basic stuff, we have Number of candidates who had taken only each of the subjects

$\begin{gathered} n(A\text{ only)=}n(A)-\lbrack n(A\cap P)+n(A\cap G)-n(A\cap P\cap G)_{}\rbrack \\ \Rightarrow152-\lbrack52+60-30\rbrack=152-82=70 \\ \\ n(G\text{ only)=}n(G)-\lbrack n(A\cap G)+n(P\cap G)-n(A\cap P\cap G)\rbrack \\ \Rightarrow136-\lbrack52+58-30\rbrack=56 \\ \\ n(P\text{ only)=}128-\lbrack58+60-30\rbrack=40 \end{gathered}$

STEP 3: Represent the values above on a Venn Diagram after getting the values

STEP 4: Find the total number of candidate that sat for the exam.

$\begin{gathered} We\text{ su}m\text{ all the numbers in the Venn diagram to get the total number of candidates that sat for the exam} \\ We\text{ have:} \\ 70+22+30+30+28+56+40=276 \end{gathered}$

Hence, the total number of candidates that sat for the exam is 276

5 0

1 year ago

Read 2 more answers