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3 years ago

8.58 divided by 6 estimate the quotient

Mathematics

2 answers:

VladimirAG [237]3 years ago

5 0

1.43 because i did 8.58/6

ololo11 [35]3 years ago

5 0

8.58/6=1.43
1.43 is about 1.4
The answer is 1.4

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On a 50 question test, Erick missed 10 questions. What is the exact grade he got on the test?

Bogdan [553]

Answer

He got an 80

Step-by-step explanation:

50 - 10 = 40

40/50 as a whole number is 80

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3 years ago

1. For a few months, Dexter recorded the amounts, in fluid ounces, of laundry detergent remaining y after he and his family wash

My name is Ann [436]

The answer would be B simply because it sounds more reasonable.

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3 years ago

Slope is given by the formula

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Answer:

Step-by-step explanation:

Answer

You never add or subtract to get the slope. That makes C and D incorrect. B is the upside down of the answer.

Answer: the answer is A slope = rise / run

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2 years ago

A 41 gram sample of a substance that's used to detect explosives has a k-value of 0.1392. Find the substance's half life, in day

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Answer:

The substance half-life is of 4.98 days.

Step-by-step explanation:

Equation for an amount of a decaying substance:

The equation for the amount of a substance that decay exponentially has the following format:

$A(t) = A(0)e^{-kt}$

In which k is the decay rate, as a decimal.

k-value of 0.1392.

This means that:

$A(t) = A(0)e^{-0.1392t}$

Find the substance's half life, in days.

This is t for which $A(t) = 0.5A(0)$ . So

$A(t) = A(0)e^{-0.1392t}$

$0.5A(0) = A(0)e^{-0.1392t}$

$e^{-0.1392t} = 0.5$

$\ln{e^{-0.1392t}} = \ln{0.5}$

$-0.1392t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.1392}$

$t = 4.98$

The substance half-life is of 4.98 days.

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3 years ago

1. Quadrilateral ABCD has vertices A(-1, 1), B(2, 3), C(6, 0) and D(3, -2). Determine using coordinate geometry whether or not t

deff fn [24]

Answer:

The Conclusion is

Diagonals AC and BD,

a. Bisect each other

b. Not Congruent

c. Not Perpendicular

Step-by-step explanation:

Given:

[]ABCD is Quadrilateral having Vertices as

A(-1, 1),

B(2, 3),

C(6, 0) and

D(3, -2).

So the Diagonal are AC and BD

To Check

The diagonals AC and BD

a. Bisect each other. B. Are congruent. C. Are perpendicular.

Solution:

For a. Bisect each other

We will use Mid Point Formula,

If The mid point of diagonals AC and BD are Same Then

Diagonal, Bisect each other,

For mid point of AC

$Mid\ point(AC)=(\dfrac{x_{1}+x_{2} }{2},\dfrac{y_{1}+y_{2} }{2})$

Substituting the coordinates of A and C we get

$Mid\ point(AC)=(\dfrac{-1+6}{2},\dfrac{1+0}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Similarly, For mid point of BD

Substituting the coordinates of B and D we get

$Mid\ point(BD)=(\dfrac{2+3}{2},\dfrac{3-2}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Therefore The Mid point of diagonals AC and BD are Same

Hence Diagonals,

a. Bisect each other

B. Are congruent

For Diagonals to be Congruent We use Distance Formula

For Diagonal AC

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting A and C we get

$l(AC) = \sqrt{((6-(-1))^{2}+(0-1)^{2} )}=\sqrt{(49+1)}=\sqrt{50}$

Similarly ,For Diagonal BD

Substituting Band D we get

$l(BD) = \sqrt{((3-2))^{2}+(-2-3)^{2} )}=\sqrt{(1+25)}=\sqrt{26}$

Therefore Diagonals Not Congruent

For C. Are perpendicular.

For Diagonals to be perpendicular we need to have the Product of slopes must be - 1

For Slope we have

$Slope(AC)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting A and C we get

$Slope(AC)=\dfrac{0-1}{6--1}\\\\Slope(AC)=\dfrac{-1}{7}$

Similarly, for BD we have

$Slope(BD)=\dfrac{-2-3}{3-2}\\\\Slope(BD)=\dfrac{-5}{1}$

The Product of slope is not -1

Hence Diagonals are Not Perpendicular.

6 0

4 years ago