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2 years ago

Can someone help me?

Mathematics

1 answer:

olga nikolaevna [1]2 years ago

7 0

The answer is -8°F. This is because it DROPPED 8 degrees, meaning it was 8°F MINUS 8°F which would equal 0°F. Hope that helps!

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Adjust the number into correct standard form. 2206×10^9

pentagon [3]

Answer:

2.306

Step-by-step explanation:

u while times the number

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2 years ago

Find the HCF and LCM of: x^3+27,2x^3-6x^2+18x,x^2-3x+9

r-ruslan [8.4K]

Answer:

hcf:3x^3-6x^2+y^2-3x+54

lcm:2(x+3)(x^2-3x+9)(x^3-3x^2+9)(y^2-3x+9)

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2 years ago

To evaluate whether a new fertilizer improves plant growth more than the old fertilizer, the fertilizer developer gives some pla

olga2289 [7]

Answer:

B) Experiment: you imposed a treatment (new fertilizer on some plants, old on others) to determine its effect.

Step-by-step explanation:

In this situation, the fertilizer developer wants to evaluate a hypothesis, not just observe, by subjecting plants to two different treatments (new and old fertilizer) in order to determine which treatment provides a more desirable effect (more plant growth).

This process configures an experiment.

The answer is alternative B)

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2 years ago

Deana bought 12 packages of batteries. Some of the packages contained 4 batteries and some contained 6 batteries. She bought a t

NARA [144]

Answer:

She bought 7 battery packs containing 6 batteries

Step-by-step explanation:

4bx5= 20 batteries

6bx7=42 batteries

42+20=62

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2 years ago

Dinosaur fossils are often dated by using an element other than carbon, like potassium-40, that has a longer half life (in this

Firlakuza [10]

The amount of substance left of a radioactive element of half life, $t_{\frac{1}{2}}$ after a time, t, is given by:

$N(t)=N_0\left( \frac{1}{2} \right)^ \frac{t}{t_{ \frac{1}{2} }}$

Given that potassium-40 has a half life of approximately 1.25 billion years.

The number of years it will take for 0.1% of potassium-40 to remain is obtained as follows:

$0.1=100\left( \frac{1}{2} \right)^ \frac{t}{1.25}} \\ \\ \Rightarrow\left( \frac{1}{2} \right)^ \frac{t}{1.25}}=0.001 \\ \\ \Rightarrow\frac{t}{1.25}\ln\left( \frac{1}{2} \right)=\ln(0.001) \\ \\ \Rightarrow \frac{t}{1.25}= \frac{\ln(0.001)}{\ln\left( \frac{1}{2} \right)} =9.966 \\ \\ t=9.966(1.25)=12.5$

Therefore, the maximum age of a fossil that we could date using 40k is 12.5 billion years.

3 0

2 years ago