All categories

2 years ago

There are over 2,700 species of snakes in the world. Over

Mathematics

1 answer:

AnnZ [28]2 years ago

7 0

Answer:

4.5

Step-by-step explanation:

2,700 ÷ 600 = 4.5

not a 100% sure just trying to help

You might be interested in

P(x) = x + 1x² – 34x + 343 d(x)= x + 9

Feliz [49]

Answer:

$x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343$

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > $\frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1$ . Go ahead and simplify.

$x=\frac{9}{d-1};\quad \:d\ne \:1$ .

Now, $\mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}$ .

$P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343$ .

Group the like terms... $1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343$ .

$\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1}$ > $1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343$ .

Now for $1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}$ .

$\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}$ .

Now for $33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}$ .

Thus we then get $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343$ .

Now we want to combine fractions. $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}$ .

$\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2$

$\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}$

$\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}$ .

Expand $81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac$ .

$-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1$ .

$\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}$

Therefore $P=\frac{-297d+378}{\left(d-1\right)^2}+343$ .

Hope this helps!

5 0

3 years ago

Here are two polygons on a grid.

NNADVOKAT [17]

Answer:

No.

Step-by-step explanation:

For polygon PQRST to be considered a scaled copy of polygon ABCDE, it means every segments of polygon ABCDE were increased proportionally by a scale factor.

The segments in polygon PQRST were not gotten using the same scale factor, hence, it is not a scaled copy of the original polygon, ABCDE.

Segment CD = 2 units, it corresponds to segment RS = 4 units. Scale factor = RS/CD = 4/2 = 2

Segment BC = 1 unit, it corresponds to segment QR = 1 unit. Scale factor = QR/BC = 1/1 = 1 units.

Varying scale factor shows polygon PQRST is not a scaled copy of polygon ABCDE.

7 0

3 years ago

In supply (and demand) problems, y is the number of items the supplier will produce (or the public will buy) if the price of the

SSSSS [86.1K]

Answer:

A) (17 ; 550)

B) $17/item

C) 550

Step-by-step explanation:

First we must calculate the intersection point of the two lines. Since in that point y has the same value in both equations, we can obtain x by equalling the two equations and then using that value for obtaining y:

$5x+465=-6x+652\\5x+6x=652-465\\11x=187\\x=\frac{187}{11}\\x=17$

So the value of x in the intersection point is 17. We now use this value with either one of the equations to obtain y. Let's use the supply equation:

$y=5*17+465\\y=550$

So the intersection point is (17 ; 550)

Supply and demand are in equilibrium when the amount of items on supply are the same as the ones on demand. That is the point were the two lines intersect, which means the selling price is the x coordinate and the amount of items is the y coordinate, so that is a selling price of $17/item with a number of items of 550.

3 0

2 years ago

Expalin what the vertical line test is and it used ?

katovenus [111]

Answer:

You use the vertical line test to see if a function is relation is a function

Step-by-step explanation:

A function is a relation with only one output for every input. That means that if the x value and two y values, it is not a function.

5 0

3 years ago

Read 2 more answers

Tan30°+cos30°÷tan30°×cos30°

jonny [76]

Answer:

{1/√3+1/2} ÷{1/√3× 1/2}

(2+√3)/2√3÷(1/2√3)
2+√3+1 /2√3
(2+1 )+√3 /2√3
3 +√3 /2√3

4 0

3 years ago