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3 years ago

The temperature was -5 degrees and then fell 8 degrees. What is the temperature now?

Mathematics

2 answers:

Schach [20]3 years ago

6 0

Answer:

-13 degrees tell me if it's right

forsale [732]3 years ago

3 0

Answer:

-13 is the answer

Step-by-step explanation:

because -5-8= -13

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La altura de una portería de futbol reglamentaria es de 2.4m y la distancia desde el punto de penalti hasta la raya de gol es de

mihalych1998 [28]

Answer:

11.06 meters

Step-by-step explanation:

Given that

The height of a regulation soccer goal is 2.4 meters

And, the distance is 10.8 meters

We need to find out the hypothesis

Here we applied the pythagoreans theorem

So,

= √ (2.4m) ^2 + (10.8 m) ^2

= √5.76 m^2 + 116.64m^2

h = 11.06 m

Hence, the hypothesis is 11.05 meters

3 0

3 years ago

HELP! Evaluate. 45 + 22 ÷ 11 • {[(60 – 15) • 60] + 25}

siniylev [52]

Step-by-step explanation: [(60 – 15) • 60] + 25} = 45.60 +25

=2725

45 + 22 ÷ 11. 2725

45 + 2 . 2725

45+ 5450

=5495

4 0

3 years ago

One of the tallest living trees today is the Mendocino Tree found at

Kisachek [45]

Answer:

the height in scientific notation is 1.12 x 10^4 :)

4 0

3 years ago

Help with this question

Alex787 [66]

Step-by-step explanation:

Hey there!

Given;

-6(2x-9) + (7-6x) = 0

Simplify it.

-12x + 54 + 7 - 6x = 0

-18x +61 = 0

Therefore, answer is option B.

Hope it helps!

6 0

3 years ago

Read 2 more answers

A sample of Iron-59 contains 50 mg. One hundred days later, the sample contains 10.75 mg. What is the half-life of iron-59, in d

exis [7]

Answer:

$t_{1/2}=45.09 d$

Step-by-step explanation:

We can use the decay equation:

$M=M_{0}e^{-\lambda t}$

M(0) is the initial mass
M is the mass after t (1000 days) time
λ is the decay constant

But:

$\lambda = ln(2)/t_{1/2}$

t(1/2) is the half-life.

So, we can rewrite the initial equation:

$M=M_{0}e^{-\frac{ln(2)}{t_{1/2}}t}$

Now, we just need to solve it for t(1/2):

$ln(\frac{M}{M_{0}})=-\frac{ln(2)}{t_{1/2}}t$

$t_{1/2}=-\frac{ln(2)}{ln(\frac{M}{M_{0}})}t$

$t_{1/2}=-\frac{ln(2)}{ln(\frac{10.75}{50})}100$

$t_{1/2}=45.09 d$

I hope it helps you!

5 0

3 years ago