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professor190 [17]
3 years ago
8

These numbers are either whole numbers, or the negatives of whole numbers. They are called ______. Example: ... -4, -3, -2, -1,

0, 1, 2, 3, 4 ...
Mathematics
2 answers:
ioda3 years ago
8 0
They are called integers

slega [8]3 years ago
7 0
Whole numbers can also be called integers. Integers can only be whole numbers and nothing else, so they cannot include decimals or fractions.
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Solve the equation 3x + 4 = 5x - 8​
CaHeK987 [17]

Answer:

x = 6

Step-by-step explanation:

3x + 4 = 5x - 8

==> add 8 to both sides

in doing so we get 3x + 4 + 8 = 5x - 8 + 8

the -8 and the +8 cancels out and 4 + 8 = 12

we are left with 3x + 12 = 5x

==> subtract 3x from both sides

in doing so we get 3x - 3x + 12 = 5x - 3x

the 3x and -3x cancel out and 5x - 3x = 2x

we are left with 12 = 2x

==> divide both sides by 2

in doing so we get 12/2 which equals 6 and 2x/2x which leaves us with

6 = x

3 0
2 years ago
Read 2 more answers
Hi, does anyone know the answer to this question? I’m bad at geometry and I’m struggling to answer it.
arsen [322]

Answer:

QT = 16

Step-by-step explanation:

ΔQRS ~ ΔQRT

In similar triangles, corresponding angles are in same ratio.

\frac{QS}{QR}=\frac{QR}{QT}\\\\\frac{25}{20}=\frac{20}{QT}

Cross multiply,

QT * 25 = 20 * 20

QT =\frac{20*20}{25}\\\\QT = 4*4\\\\QT = 16

3 0
3 years ago
Estimate the following quantities without using a calculator. Then find a more precise result, using a calculator if necessary.
zubka84 [21]

Explanation:

a. It is so easy to double a number that no approximation is necessary.

  31,000 × 200 = 3.1×10⁴ × 2×10² = 6.2×10⁶ exactly

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b. 6.2×10³ × 5.2×10⁶ ≈ 6×5×10⁹ = 3×10¹⁰ approximately

The approximation can be refined a bit by taking the ".2" into account:

  6.2×10³ × 5.2×10⁶ ≈ (6×5 + .2×(6+5))×10⁹

  = 32.2×10⁹ = 3.22×10¹⁰ approximately

Actual product: 3.224×10¹⁰.

For most purposes, the approximation is an adequate approximation, as it it within 10% of the actual value.

__

c. 9×10⁶ -2.3×10⁴ ≈ 9×10⁶ approximately

A better approximation is to actually subtract an approximation of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.02)×10⁶ ≈ 8.98×10⁶ approximately

The actual value uses all of the digits of the smaller number:

  9×10⁶ -2.3×10⁴ ≈ (9 - 0.023)×10⁶ = 8.977×10⁶ exactly

As in part B, either approximation is adequate for most purposes, as the difference from the actual value is less than .3%.

_____

The accuracy required of an approximation, hence the work you expend improving accuracy, should depend on the need in the final application of the number. Often, approximations are used for budget or resource planning purposes where some "slop" is allowed or even expected.

They can also be used in engineering applications, where the error needs to be on the side of more safety (rather than less).

7 0
3 years ago
Consider two quantities u and v that are related by the expression vp=cuq, where c is a constant. the exponents p and q are not
Mazyrski [523]

Answer:

y = (q/p)x +log(c)/p

Step-by-step explanation:

v^{p}=cu^{q}\qquad\text{original equation}\\\\p\log{(v)}=\log{(c)}+q\log{(u)}\qquad\text{take the log}\\\\py=log{(c)}+qx\qquad\text{substitute x and y}\\\\y=\dfrac{\log{(c)}+qx}{p}

3 0
3 years ago
Find the midpoint of a line segment with endpoints at (-5,7) and (13, 13).
SIZIF [17.4K]

Answer:

About 7

Step-by-step explanation:

4 0
3 years ago
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