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posledela
3 years ago
5

I have a d in my class but i need it caught up to a b or an a asap so will anyone help me please?

Mathematics
2 answers:
liraira [26]3 years ago
8 0
I’m not sure but I really wish you luck in getting your grade up!
anyanavicka [17]3 years ago
7 0

Answer:

Hey man, it wont work :(

Step-by-step explanation:

nobody will try to help you. Ive been there. GL trying to do it on your own.

Word of advice, Do it next time.

Might be boring or nor fun but its easier in the long run.

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Help plzzzzzzzzzzzzzz ​
UNO [17]

Answer:

II only ⇒ QR ≅ SP

Step-by-step explanation:

It definitely can’t be I or III becasue SR is smaller than QR, and PR is longer than QR. So both of those lines are not congruent to QR because they have different lengths.

7 0
3 years ago
Confused on this help plss
ValentinkaMS [17]

Answer:

of iv

+

2

(

4

−

)

=

2

+

6

−

3

x+2({\color{#c92786}{4-x}})=2x+6-3x

x+2(4−x)=2x+6−3x

+

2

(

−

+

4

)

=

2

+

6

−

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Step-by-step explanation:

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8 0
2 years ago
Why might you expect most absolute value equations to have two solutions? Why not three or four?
Travka [436]

Answer:

If the absolute value expression is not equal to zero, the expression inside an absolute value can be either positive or negative. So, there can be at most two solutions. Looking at this graphically, an absolute value graph can intersect a horizontal line at most two times.

3 0
3 years ago
The owner of a new coffee shop is keeping track of how much each customer spends (in dollars). A random number table was used to
frosja888 [35]
I expect the mean of the sample to be 5.00
4 0
2 years ago
Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).
lions [1.4K]

Check the picture below.  so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}

(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}

6 0
2 years ago
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