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larisa86 [58]
3 years ago
5

The weight of an object on a particular scale is 142.2 lbs the measured weight may vary from the actual weight by at most 0.3 lb

s. what is the range of actual weights of the object
Mathematics
1 answer:
algol133 years ago
3 0

Answer:

144.9 <= x <= 145.5

Step-by-step explanation:

145.2 + 3 = 145.5

145.2 - 3 = 144.9

144.9 <= x <= 145.5

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A stadium has seating accommodation for 3800 people. At a cricket match all seats were taken. If there were 5609 people attendin
umka2103 [35]

Answer:

Step-by-step explanation:

Those standing is the same thing as people who didn't have seats.

Those standing = the total number of people attending - those seated.

Those standing = 5609 - 3800

Those standing = 1809

6 0
3 years ago
Round the decimal to the nearest hundredths 983.625
r-ruslan [8.4K]
983.625
9 = hundreds
8 = tens
3 = ones
6 = tenths
2 = hundredths
5 = thousandths
You want to round to the nearest hundredths
Look at the number on the right, if it's bigger than 5, round up, if it's 4 or lower, round down.
In this case, the number to the right is a 5, so you round the 2 (hundredths) up one so:
<span>983.63</span>
5 0
3 years ago
Which statement is true?
Greeley [361]

Answer:

Step-by-step explanation:

A The axis of symmetry of a parabola will always pass through the vertex.   TRUE!

B The axis of symmetry of a parabola will sometimes pass through the vertex.  FALSE!

C The axis of symmetry of a parabola will never pass through the vertex.  FALSE

4 0
3 years ago
2+cotA=1 homework help
Alex Ar [27]
2+\cot A=1\implies \cot A=-1\implies \tan A=-1

This happens whenever A=\dfrac{3\pi}4 or A=\dfrac{7\pi}4. More generally, \tan A=-1 whenever you start with one of these angles and add any multiple of \pi, so the general solution would be A=\dfrac{3\pi}4+n\pi, where n is any integer. (Notice that when n=1, you end up with \dfrac{7\pi}4.)
3 0
4 years ago
Does anyone know how to do this and if so can you please help me and explain how to do it, it’ll be appreciated thank you
dalvyx [7]

Answer:

13) (5x)^{-\frac{5}{4} ⇒ \frac{1}{\sqrt[4]{(5x)^5}}

15) (10n)^{\frac{3}{2} ⇒ \sqrt{(10n)^3}

Step-by-step explanation:

Given expression:

13) (5x)^{-\frac{5}{4}

15) (10n)^{\frac{3}{2}

Write the expressions in radical form.

Solution:

For an expression with exponents as fraction like

(x)^{\frac{m}{n}

the numerator m represents the power it is raised to and the denominator n represents the nth root of the expression.

For an expression with exponents as negative  fraction like

(x)^{-\frac{m}{n}

We take the reciprocal of the term by rule for negative exponents.

So it is written as:

\frac{1}{(x)^{\frac{m}{n}}}

using the above properties we can write the given expressions in radical form.

13) (5x)^{-\frac{5}{4}

⇒ \frac{1}{(5x)^{\frac{5}{4}}}   [Using rule of negative exponents]

⇒ \frac{1}{\sqrt[4]{(5x)^5}}    [writing in radical form]

15) (10n)^{\frac{3}{2}

⇒ \sqrt{(10n)^3}     [Since 2nd root is given as \sqrt{} in radical form]

3 0
3 years ago
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