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1 year ago

{[(8-3)x2]+[5x6)-5]} divided by five

2 answers:

6 0

{[(8 - 3) * 2] + [(5 * 6) - 5]} / 5 =
= {[2 * 8 + 2 * (-3)] + 30 - 5} / 5 =
= [(16 - 6) + 25] / 5 =
= (10 + 25) / 5 =
= 35 / 5 =
= 7

Brackets make it look complex but it's not that bad at all :)

nignag [31]1 year ago

3 0

Answer:

The answer is 7.

Step-by-step explanation:

First we will solve this as per pemdas.

${[(8-3)\times2]+[(5\times6)-5]}$

= ${[5\times2]+[(5\times6)-5]}$

=> ${[10]+[(30)-5]}$

=> ${[10]+[25]}$

= 35

Now 35 divided by 5 = 7

$\frac{35}{5}=7$

Therefore, the answer is 7.

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2 years ago

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2 years ago

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Let a and b be real numbers where a=/b=/c=/0 which of the following functions could represent the graph below?

Tresset [83]

Answer: The second option; y = (x - a)^2*(x-b)^4

Step-by-step explanation:

Ok, we have that a and b are real numbers different than zero.

In the graph, we can see that the line touches the x-axis in two values. Now, if we would have an equation like:

y = x*(x - a)^3*(x - b)^3

then when x = 0 we would have:

y = 0*(0-a)^3*(0-b)^3 = 0

But in the graph, we can see that when x = 0, the value of y is different than zero, so we can discard options 1 and 3.

So the remaining options are:

y = (x - a)^2*(x-b)^4

y = (x - a)^5*(x - b)

Now, another thing you can see in the graph is that it is always positive.

Particularly the second option allows negative values for y because it has odd powers, then we can also discard this option.

(For example, if x > a and x < b we would have a negative value for y)

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2 years ago

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

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2 years ago