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3 years ago

ASAP I really need help please!

Mathematics

2 answers:

slava [35]3 years ago

7 0

Answer:

I can't see the fine details of the image displaying the weights of the bags but I would assume the heaviest bag is 2(3/4) pounds, being Bag C. The lightest bag is most likely Bag B, being 2(3/8) pounds.

VikaD [51]3 years ago

3 0

Answer:

I can't read the fraction for B

but 2 3/4 is bigger than 2 1/2

You might be interested in

Can someone please explain AAS SAS SSS ASA CPCTC

Lostsunrise [7]

The Angle Angle Side postulate (often abbreviated as AAS) states that if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent.

The Side Angle Side postulate (often abbreviated as SAS) states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

SSS stands for "side, side, side" and means that we have two triangles with all three sides equal. For example: is congruent to: (See Solving SSS Triangles to find out more) If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.

ASA stands for "angle, side, angle" and means that we have two triangles where we know two angles and the included side are equal. If two angles and the included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent.

CPCTC is an acronym for corresponding parts of congruent triangles are congruent. CPCTC is commonly used at or near the end of a proof which asks the student to show that two angles or two sides are congruent. ... Corresponding means they're in the same position in the 2 triangles.

6 0

3 years ago

Find the angle measures. Justify your responses/

Musya8 [376]

Answer:

m6 = 117 degrees

m7 = 63 degrees

Step-by-step explanation:

I just asked my friend 0_0

6 0

3 years ago

Read 2 more answers

Jenny works in a picture framing shop. She has a function f(x) equals X +2 that find the size of a square frame, given a picture

Iteru [2.4K]

Given:

The function for size of a square frame is

$f(x)=x+2$

where, x is the side length of the picture.

The function for the price in dollars for the frame is

$p(x)=3x$

To find:

The single function for the price of a picture with an edge length of x.

Solution:

We know that, for a picture with an edge length of x.

Size of a square frame = f(x)

Price in dollars for the frame = p(x)

Single function for the price of a picture with an edge length of x is

$(p\circ f)(x)=p(f(x))$

$(p\circ f)(x)=p(x+2)$ $[\because f(x)=x+2]$

$(p\circ f)(x)=3(x+2)$ $[\because p(x)=3x]$

Let the name of this function is c(x). So,

$c(x)=3(x+2)$

Therefore, the required function is $c(x)=3(x+2)$ .

8 0

3 years ago

Can someone explain if I am wrong?

snow_lady [41]

Answer:

Step-by-step explanation:

A. (b - 10)(b + 10) = 180

$b^{2}$ -100 = 180

$b^{2}$ = 280

b = $\sqrt{280}$

b = 2 $\sqrt{70}$

Your A is correct.

B. 2 $\sqrt{70}$ - 10 = c

c = 6.733200531 = 6.73 Try rounding to 3 significant figures.

C. 2 $\sqrt{70}$ + 10 = d

d = 26.73320053 = 26.7

You are correct in all 3 answers. Perhaps you should try rounding to 3 significant figures for your c and d values.

I hope this helps!! ^-^

8 0

3 years ago

Read 2 more answers

Compute the sum:

Nady [450]

You could use perturbation method to calculate this sum. Let's start from:

$S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}$

On the other hand, we have:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}$

So from (1) and (2) we have:

$\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\
S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\
(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}$

Now, let's try to calculate sum $\sum\limits_{k=0}^{n}k\cdot k!$ , but this time we use perturbation method.

$S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\
\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\
$

but:

$S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=
\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=
\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\
\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}$

When we join both equation there will be:

$\begin{cases}S_{n+1}=S_n+(n+1)(n+1)!\\\\S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\end{cases}\\\\\\
S_n+(n+1)(n+1)!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\\\
\sum\limits_{k=0}^{n}(k^2+1)k!=S_n-2S_n+(n+1)(n+1)!=(n+1)(n+1)!-S_n=\\\\\\=
(n+1)(n+1)!-\sum\limits_{k=0}^nk\cdot k!\stackrel{(\star)}{=}(n+1)(n+1)!-[(n+1)!-1]=\\\\\\=(n+1)(n+1)!-(n+1)!+1=(n+1)!\cdot[n+1-1]+1=\\\\\\=
n(n+1)!+1$

So the answer is:

$\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}$

Sorry for my bad english, but i hope it won't be a big problem :)

8 0

4 years ago