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anyanavicka [17]
4 years ago
7

What is the smallest odd number with four different prime factors?

Mathematics
2 answers:
nevsk [136]4 years ago
8 0
Obviously the smallest odd number we are looking for will have the SMALLEST four prime factors. 

The smallest four prime numbers are 2,3,5,7. The lowest multiple of this is 2*3*5*7 = 210 But this is an even number! 

So let's remove 2, and use the next prime number: 3,5,7,11. The lowest multiple of these is 3*5*7*11 = 1155.


torisob [31]4 years ago
5 0
The answer is 210 since the numbers are (2x3x5x7)
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Wewaii [24]

Answer:

X= -3

Step-by-step explanation:

8 0
3 years ago
A plane is 24,000 feet above sea
IrinaVladis [17]

Answer:

A. -500

Step-by-step explanation:

Find the slope using rise over run (change in y / change in x)

Find the change in the y values (height of the plane):

24,000 - 18,000

= 6,000

Find the change in the x values (miles from the airport)

(60 - 72)

= -12

Divide 6000 by -12:

6000/-12

= -500

So, the slope of the plane's path is A. -500

5 0
3 years ago
Read 2 more answers
If geometry symbol represented as small triangle with three sides. abc geometry congruent symbol represented as two small horizo
n200080 [17]

The value of x in the congruent triangles abc and dec is 1

<h3>How to determine the value x?</h3>

The question implies that the triangles  abc and dec are congruent triangles.

The congruent sides are:

ab = de

bc = ce = 4

ac = cd = 5

The congruent side ab = de implies that:

4x - 1 = x + 2

Collect like terms

4x - x = 2 + 1

Evaluate the like terms

3x = 3

Divide through by 3

x = 1

Hence, the value of x is 1

Read more about congruent triangles at:

brainly.com/question/12413243

#SPJ1

<u>Complete question</u>

Two triangles, abc and cde, share a common vertex c on a grid. in triangle abc, side ab is 4x - 1, side bc is 4, side ac is 5. in triangle cde, side cd is 5, side de is x + 2, side ce is 4. If Δabc ≅ Δdec, what is the value of x? a. x = 8 b. x = 5 c. x = 4 d. x = 1 e. x = 2

3 0
2 years ago
There are 7/8 kilograms of salt in the kitchen. Mrs. Jackson used 2/15 of salt when preparing dinner . How much salt did she use
neonofarm [45]

so since Mrs Jackson used 2/15 of 7/8, the amount she used is simply their product.


\bf \stackrel{\textit{how much is }\frac{2}{15}\textit{ of }\frac{7}{8}?}{\cfrac{7}{8}\cdot \cfrac{2}{15}}\implies \cfrac{7}{15}\cdot \cfrac{2}{8}\implies \cfrac{7}{15}\cdot \cfrac{1}{4}\implies \cfrac{7}{60}

8 0
3 years ago
For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
pav-90 [236]

By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

4 0
3 years ago
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