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4 years ago

67.000 rounded to the nearest ten thousand

Mathematics

2 answers:

tankabanditka [31]4 years ago

6 0

70.000 not really sure

ycow [4]4 years ago

3 0

70.000 When The Number Is 5 And Up You Count Up One When The Number Is 4 And Lower You Count Down One

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The base of a triangle is 4 cm greater than the

Anna35 [415]

Answer:

Step-by-step explanation:

Formula for area of a triangle:

Height x Base /2

Base (b) = h +4

Height = h

h + 4 x h /2 = 30cm

=> h +4 x h = 60

=> h+4h =60

=> 5h = 60

=> h = 12

Height = 12

Base = 12 +4 = 16

4 0

3 years ago

Write an equation for a polynomial with all the following features: The Degree is 5. The graph of the Polynomial crosses the ver

dimulka [17.4K]

Answer: f(x) = a*x^5 + b*x^3 + c*x^2 + 12.

Step-by-step explanation:

The degree is 5, so we will have a term like:

a*x^5

it crosses the vertical axis at y = 12, then we will have a constant term equal to 12.

The polynomial has 4 terms, and we already defined two, so we can invent two more, such that the exponent must be between 1 and 4

This polynomial can be something like:

f(x) = a*x^5 + b*x^3 + c*x^2 + 12.

where a, b and c are real numbers.

Has 4 terms, f(0) = 12, then it intersects the y-axis at y = 12, and the maximum exponent is 5, then the degree of f(x) is 5.

4 0

4 years ago

What is 6,000 milliters=___6 lilters

worty [1.4K]

Answer:

its a conversion you mutlply by 100

Step-by-step explanation:

8 0

3 years ago

Round to the nearst tenth

Serga [27]

Answer:

Step-by-step explanation:

4 0

3 years ago

50 POINTS!!! In rectangle ABCD, AB = 6 cm, BC = 8 cm, and DE = DF. The area of triangle DEF is one-fourth the area of rectangle

aalyn [17]

Answer:

$EF=4\sqrt{3}$

Step-by-step explanation:

In rectangle ABCD, AB = 6, BC = 8, and DE = DF.

ΔDEF is one-fourth the area of rectangle ABCD.

We want to determine the length of EF.

First, we can find the area of the rectangle. Since the length AB and width BC measures 6 by 8, the area of the rectangle is:

$A_{\text{rect}}=8(6)=48\text{ cm}^2$

The area of the triangle is 1/4 of this. Therefore:

$\displaystyle A_{\text{tri}}=\frac{1}{4}(48)=12\text{ cm}^2$

The area of a triangle is half of its base times its height. The base and height of the triangle is DE and DF. Therefore:

$\displaystyle 12=\frac{1}{2}(DE)(DF)$

Since DE = DF:

$24=DF^2$

Thus:

$DF=\sqrt{24}=\sqrt{4\cdot 6}=2\sqrt{6}=DE$

Since ABCD is a rectangle, ∠D is a right angle. Then by the Pythagorean Theorem:

$(DE)^2+(DF)^2=(EF)^2$

Therefore:

$(2\sqrt6)^2+(2\sqrt6)^2=EF^2$

Square:

$24+24=EF^2$

Add:

$EF^2=48$

And finally, we can take the square root of both sides:

$EF=\sqrt{48}=\sqrt{16\cdot 3}=4\sqrt{3}$

6 0

3 years ago

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