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sladkih [1.3K]
3 years ago
7

PLLZZ HELP ME OUT WITH THIS!!!!

Mathematics
2 answers:
Anna007 [38]3 years ago
5 0

Answer:

for 1. -30 - 12 = -42

Step-by-step explanation:

Andrei [34K]3 years ago
4 0
She is at -30 feet then dives down 12 more

30+12=42
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Solve the quadratic equations below.<br> 3x2 + 22x + 35 = 0
mestny [16]

Step-by-step explanation:

3x² + 22x + 35 = 0

(3x + 7)(x + 5) = 0

3x + 7 = 0

x = -7/3

or x+5 = 0

x = -5

3 0
3 years ago
The sum of two numbers is 21. The larger number is 6 less than twice the smaller number. Find the two numbers.
Semenov [28]
The two numbers are 9 and 12.
6 0
3 years ago
0.10 x 37.7 explain please
kodGreya [7K]

Answer: The answer is 3.77.

Explanation: Since 0.10 is 10 times smaller than 1, we know that if we divide 37.7 by 10, we will get the same answer as 0.10 x 37.7.

5 0
2 years ago
(UPE) Terremotos são eventos naturais que não têm relação com eventos climáticos extremos, mas podem ter consequências ambientai
Alex

Answer:

E=10^{18} Joules

Step-by-step explanation:

1) Olá! Adicionando dados faltantes ao enunciado, como a fórmula dada: M=\frac{2}{3}log(\frac{E}{E_{0}})\\ e corrigindo a quantidade de Energia inicial: 10^{4,5} Vamos seguir aplicando a fórmula, lembrando das propriedades de logaritmo.

M=\frac{2}{3}log(\frac{E}{E_{0}})\\9=\frac{2}{3}log(\frac{E}{10^{4,5}})*3\\27=2log(\frac{E}{10^{4.5}})\\log(\frac{E}{10^{4.5}})^{2}=27\\(\frac{E^{2}}{10^{9}})=10^{27}\\E^{2}=10^{27}*10^{9}\Rightarrow E=\sqrt{10^{36}}\rightarrow E=10^{18} Joules

6 0
3 years ago
(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2
Montano1993 [528]

Answer:

\dfrac{m^3}{16p^2q^2}

Step-by-step explanation:

Given:

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}

1.

m^{-1}=\dfrac{1}{m}

2.

q^0=1

3.

2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}

4.

(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}

5.

m^{-1}=\dfrac{1}{m}

6.

2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}

7.

\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}

8.

(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}

8 0
3 years ago
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