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Lera25 [3.4K]
3 years ago
5

I need help on the values on these questions?

Mathematics
1 answer:
Hitman42 [59]3 years ago
5 0

Answer:

Angle 6 is 135 degrees

Step-by-step explanation:

Reason why is because look at 45. (9x+14 is corresponding to that so they must have the same values.

meaninng 180-45 gets you 135. So, 135 is the measure of angle 6

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Marian went shopping and bought clothes for $66.17 and books for $44.98. She then had a meal at the mall for $20.15. Which is th
scZoUnD [109]

Answer:

Either 130 or 131 would be acceptable.

Step-by-step explanation:

When we estimate, we do not use exact values.

66.17 is close to 66

44.98 is close to 45

20.15 is close to 20

           

Add the estimates together: 66+ 45+ 20 = 131

We could also estimate 66.17 to 65 which would give us an estimate of 130




6 0
3 years ago
Read 2 more answers
If the factors of quadratic function g are (x − 7) and (x + 3), what are the zeros of function g?
MArishka [77]

Answer: B. -3 and 7

Step-by-step explanation:

To find the zeros, just move the constants to the other side,

x - 7 ⇒ x = 7

x + 3 ⇒ x = -3

If the number is moved to the other side its sign becomes opposite

Hope it helps!

8 0
2 years ago
Every 24 hours, the earth rotates 360° about its axis. How
Usimov [2.4K]

360 degrees / 24 hours = 15 degrees per hour


150 degrees / 15 degrees per hour = 10 hours


Answer: 10 hours

8 0
2 years ago
How to solve logarithmic equations as such
Serga [27]

\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\\\ \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\textit{we'll use this one}}{a^{log_a x}=x} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}

\bf \log_2(x-1)=\log_8(x^3-2x^2-2x+5) \\\\\\ \log_2(x-1)=\log_{2^3}(x^3-2x^2-2x+5) \\\\\\ \log_{2^3}(x^3-2x^2-2x+5)=\log_2(x-1) \\\\\\ \stackrel{\textit{writing this in exponential notation}}{(2^3)^{\log_2(x-1)}=x^3-2x^2-2x+5}\implies (2)^{3\log_2(x-1)}=x^3-2x^2-2x+5

\bf (2)^{\log_2[(x-1)^3]}=x^3-2x^2-2x+5\implies \stackrel{\textit{using the cancellation rule}}{(x-1)^3=x^3-2x^2-2x+5} \\\\\\ \stackrel{\textit{expanding the left-side}}{x^3-3x^2+3x-1}=x^3-2x^2-2x+5\implies 0=x^2-5x+6 \\\\\\ 0=(x-3)(x-2)\implies x= \begin{cases} 3\\ 2 \end{cases}

5 0
3 years ago
Help me pls don’t put random
Vladimir [108]
I think it’s A
Sorry if I’m wrong
4 0
3 years ago
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