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3 years ago

How many shares of Lombard Incorporated traded on the 17th?

Mathematics

1 answer:

Naily [24]3 years ago

8 0

Answer:

Step-by-step explanation:

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A:Consecutive interior angles

kykrilka [37]

Answer:

D: corresponding angle

hope it helps.

5 0

3 years ago

Read 2 more answers

What is it quick. And drop y’all Insta ask more

Tcecarenko [31]

Answer:

225 m2

Step-by-step explanation:

area= side*side

= 15x15

225 m2

4 0

3 years ago

Taylor ran 2 1/4 miles and walked 2 4/5 miles. How far did she run and walk

irinina [24]

Okay. So, we should convert fractions to the least common denominator, which is 20. That would be 2 5/20 and 2 16/20. Add both of those numbers up to get 5 1/20. Taylor ran and walked 5 1/20 miles.

5 0

3 years ago

If 5 + 20 times 2 Superscript 2 minus 3 x Baseline = 10 times 2 Superscript negative 2 x Baseline + 5, what is the value of x? –

sp2606 [1]

Answer:

<h3>Option D) 3 is correct</h3><h3>Therefore the value of x is 3</h3>

Step-by-step explanation:

Given equation is $5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5$

<h3>To find the value of x :</h3>

First solving the given equation we have,

$5+20\times (2)^{2-3x}=10\times (2)^{-2x}+5$

$5+20\times (2)^{2-3x}-5=10\times (2)^{-2x}+5-5$

$20\times (2)^{2-3x}=10\times (2)^{-2x}$

$20\times (2)^2.(2)^{-3x}=10\times (2)^{-2x}$

$20\times 4.(2)^{-3x}=10\times (2)^{-2x}$

$80(2)^{-3x}=10\times (2)^{-2x}$

$\frac{80}{10}(2)^{-3x}=\frac{10\times (2)^{-2x}}{10}$

$8(2)^{-3x}=(2)^{-2x}$

$\frac{(2)^{-3x}}{(2)^{-2x}}=\frac{1}{8}$

$(2)^{-3x}.(2)^{2x}=\frac{1}{2^3}$ ( by using the property $\frac{1}{a^{-m}}=a^m$ )

$2^{-3x+2x}=\frac{1}{2^3}$ ( by using the property $a^m.a^n=a^{m+n}$ )

$2^{-x}=\frac{1}{2^3}$

$\frac{1}{2^x}=\frac{1}{2^3}$ ( by using the property $a^{-m}=\frac{1}{a^m}$ )

Since bases are same so powers are same

Therefore we can equate the powers we get x=3

<h3>Therefore the value of x is 3</h3><h3>Option D) 3 is correct</h3>

5 0

3 years ago

Read 2 more answers

I’m Really lost if I could get a answer it would be greatly appreciated

Nataly [62]

<h3>Answers:</h3>

$f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}$

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Work Shown:

Part 1

$f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\$

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

$g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\$

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out $\left(\sqrt{x-1}+5\right)^2$

6 0

3 years ago