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svp [43]
3 years ago
15

Please help me with this

Mathematics
1 answer:
777dan777 [17]3 years ago
6 0

Step-by-step explanation:

3)

(2x-3)(2x+3)

2x(2x+3)-3(2x+3)

4x²+6x²-6x-9

8x²-6x-9

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Look for Patterns Find the missing numbers in the following pattern. 1,3,9, .81 .​
ehidna [41]

Answer:

27 after 81 is 243

Step-by-step explanation:

The pattern is that you multiply 3 each time.

5 0
3 years ago
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What is the correct answer to this ?​
marin [14]

Answer:

no

Step-by-step explanation:

its a scalene triangle

8 0
3 years ago
PLEASE HELP!!! The amount of polonium-210 remaining, P(t), after t days in a sample can be modeled by the exponential function P
exis [7]

Answer:

P(t)=100(0.994)^t

Step-by-step explanation:

A calculator shows the value of e^{-.006} \approx 0.994 meaning that each day, about 99.4% of the sample remains.

The formula P(t)=100e^{-0.006t} = 100(e^{-0.006})^t=100(0.994)^t

Evaluate this at <em>t</em> = 16 to get 91 grams remain.

3 0
3 years ago
mandy,jermey and lily went to an amusement park during their summer vacation mandy spent 16.25$ at the amusement park jermey spe
erastovalidia [21]
Mandy: $16.25
Jermey: $16.25 + $3.40 = $19.65
Lily: $19.65 x 2 = $39.30
For Lily you would take the total of Jermey and times it by 2.
4 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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