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3 years ago

Simplify 2-5-(12÷4-1)=

1 answer:

5 0

Use bidmas
so in order of what to do its Brackets, Indices, Division, Multiplication, Add, Subtract
hence you as (12/4 - 1) is in brackets, and 12/4 is division, you do 12/4 = 3
3 - 1 = 2
2 - 5 - 2 = - 5

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Does this graph show a function? Explain how you know.

beks73 [17]

Answer:

No, the graph fails the vertical line test

Step-by-step explanation:

To determine if the graph is a function, we can use the vertical line test.

Use a vertical line, if the vertical passes through two or more points, the graph is not a function

Looking at the y axis ( which is a vertical line), it passes through two points

This means the graph is not a function

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2 years ago

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krek1111 [17]

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3 years ago

Read 2 more answers

If a=2b^3 and b= -1/2c ^-2,express a in terms of c.

ElenaW [278]

Answer:

Part 1) $a=-\frac{1}{4c^6}$

Part 2) $a=-\frac{1}{4c^{-6}}$

Step-by-step explanation:

Part 1) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2}c^{-2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{1}{2}c^{-2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{1}{2}c^{-2})^{3}=2(-\frac{1}{2})^3(c^{-2})^{3}=2(-\frac{1}{8})(c^{-6})=-\frac{1}{4c^6}$

therefore

$a=-\frac{1}{4c^6}$

Part 2) we have

$a=2b^{3}$ ----> equation A

$b=-\frac{1}{2c^{-2}}=-\frac{c^{2}}{2}$ ----> equation B

substitute equation B in equation A

$a=2(-\frac{c^{2}}{2})^{3}$

Applying property of exponents

$(x^{m})^{n}=x^{m*n}$

$x^{-m} =\frac{1}{x^{m}}$

$a=2(-\frac{c^{2}}{2})^{3}=2(-\frac{c^{6}}{8})$

simplify

$a=-\frac{c^{6}}{4}$

therefore

$a=-\frac{1}{4c^{-6}}$

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3 years ago

Whoever answers I’ll mark brainliest

Andre45 [30]

Answer:

x is 60⁰ because 180 - 40 is 120 divide it by 2 because the bottem angles are equal and u get 60

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3 years ago

Determine the values of the constants b and c so that the function given below is differentiable. f(x)={2xbx2+cxx≤1x>1

Lera25 [3.4K]

Assuming the function is

$f(x)=\begin{cases}2x&\text{for }x\le1\\bx^2+cx&\text{for }x>1\end{cases}$

For $f(x)$ to be differentiable, it necessarily has to be continuous. For this condition to be met, we need

$\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)$
$\iff\displaystyle\lim_{x\to1}2x=\lim_{x\to1}(bx^2+cx)$
$\iff2=b+c$

For the derivative to exist, the one-sided limits of the derivative must also exist and be equal. We have

$f'(x)=\begin{cases}2&\text{for }x1\end{cases}$

$\displaystyle\lim_{x\to1^-}2=\lim_{x\to1^+}(2bx+c)$
$\iff2=2b+c$

Now we solve for $b$ and $c$ :

$\begin{cases}b+c=2\\2b+c=2\end{cases}\implies b=0,c=2$

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3 years ago