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kirill115 [55]
3 years ago
6

Shane's money is 3:11 if Isabella has 33 dollars how much do shane and Isabella have together

Mathematics
1 answer:
grin007 [14]3 years ago
3 0
3:11
S:I
?:33
So if Isabella has 33 dollars that means they did 11x3=33 so to find Shanes you would do 3x3=9 then add the two together which is $42 hope this helps
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Alex bought a T-shirt for $24.25. This week the T-shirt costs $5.95 less. Write and evaluate an addition expression to find this
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Answer: 24.25 + (-5.95)

Explanation:

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3 years ago
F(x)=-2x+1; {-2,0,2,4,6} Find the range of the function for the given domain.
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Answer:

For the domain indicated, the range is  {-11, -7, -3, 1, 5}

Step-by-step explanation:

Given the indicated domain of f(x) you have to replace the values in the function to find the range (the "y" values of the function).

f(-2) = 5

f(0) = 1

f(2) = -3

f(4) = -7

f(6) = -11.

So the range for this domain is {-11, -7, -3, 1, 5}

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3 years ago
Which formula is easier to use. Finding the area of a trapezoid or circle? Explain in detail with a minimum of 4 sentence
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Step-by-step explanation:

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Circle Formula: πr*2

Solving a circle requires less steps than solving for a trapezoid would, so in my opinion, finding the area of a circle is easier than a trapezoid.

Hope this helps!

8 0
3 years ago
The value of the x-intercept for the graph of 4x-5y=40 is
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\hbox{x-intercept - y=0}: \\&#10;4x-5 \times 0=40 \\&#10;4x=40 \\&#10;x=\frac{40}{4} \\&#10;\boxed{x=10}
8 0
3 years ago
I need help asap pls and thank you ;)
olga55 [171]

Answer:

\text{Length of AB is }\frac{ah}{a+h}

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP     (∵ corresponding angles)

∠NDC=∠NPM     (∵ corresponding angles)

By AA similarity rule,  ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\

KA(\frac{h}{x}-1)=AP

ND(\frac{h}{x}-1)=PD

Adding above two, we get

(KA+ND)(\frac{h}{x}-1)=(AP+PD)

⇒ (KN-AD)=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x}{(\frac{h}{x}-1)}

⇒ a-x=\frac{x^2}{h-x}

⇒ x^2=ah-ax-xh+x^2

⇒ x(h+a)=ah

⇒ x=\frac{ah}{a+h}

3 0
3 years ago
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