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2 years ago

Can the product of a whole number and a decimal number less than 1 ever be greater than a whole number

1 answer:

7 0

No. The product of a whole number and a decimal number less than 1 will be greater than the whole number multiplied into. For this theorem to be proven. Let us state the mathematical expression into numbers such that 
1. N x 0.1 = N/0.1 < N 
2. 1 x 0.5 = 0.5 
3. 2 x 0.1 = 0.2 
4. 100 x 0.55 = 55

These three examples and stances then suggest the claim that the product is not equal to the whole number used in the equation.

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Match the given angle pair to their relationship using the figure beloe

hoa [83]

Answer:

Vertical: 1

corresponding: 3

alternate interior: 2

alternate exterior: 4

6 0

2 years ago

Help please :) ..................

brilliants [131]

Answer:

1587.6 m²

Step-by-step explanation:

$A=\frac{1}{2} bh$ where b is the base length and h is the height (b and h on the triangle will always be perpendicular to each other)

Plug in 75.6 m as b and 42 m as h

$A=\frac{1}{2} (75.6)(42)\\A=\frac{1}{2}( 3175.2)\\A=1587.6$

I hope this helps!

3 0

2 years ago

I need helpabdjbajdbajdb

gizmo_the_mogwai [7]

Answer:

c=10

Step-by-step explanation:

pythagreom theroem

c (hypotenuse) = $\sqrt{a^{2} +b^{2} }$

plug in values for a and b

c= $\sqrt{6^{2} +8^{2} }$

c= $\sqrt{36+64\\}$

c= $\sqrt{100\\}$

c=10

7 0

2 years ago

Read 2 more answers

Solve for the variable c in this equation a(c + b) = d

noname [10]

Multiply everything in the parenthesis by a.

ac + ab = d

Subtract ab from both sides.

ac = d - ab

Divide a on both sides.

c = d - ab / a

Hope this helps!

3 0

2 years ago

a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar

Margarita [4]

A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem:

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus,

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$ (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.

So, the total surface area = base area + lateral area = $24 \sqrt{3}+72$ cm^2

Answer: $24 \sqrt{3}+72$ cm^2

4 0

2 years ago