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Gelneren [198K]

4 years ago

14

In a class of 500 students a hundred and sixty have 320 have brown eyes and 20 green in a class of 25 students how many would ha

ve blue eyes

1 answer:

saveliy_v [14]4 years ago

4 0

Is your question right because you said 500 students a hundred and sixty have 320 have brown eyes?

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Louisa is trying to read her grocery receipt, but it is blurry. The sales tax rate was 6.25%. She pays tax only on the items lab

Artyom0805 [142]

Tax on canned tuna : None
Tax on paper towels : 6.89*6.25/100=0.430625~$0.43
Tax on straws : 0.99*6.25/100=0.061875~$0.06
Tax on batteries : 5.67*6.25/100=0.35475~$0.35

Total taxes : 0.43+0.06+0.35=$0.84

3 0

3 years ago

Find the value of x. Plz hurry

WITCHER [35]

Answer:

x = 12

Step-by-step explanation:

both (x+5) and (2x-7) are corresponding angles formed by an intersection between a transversal and 2 parallel lines.

hence they are equal in value

x+5 = 2x-7

x - 2x = -7 - 5

-x = -12

x = 12

5 0

4 years ago

Determine whether 5x^3 – 25y is a polynomial. If so, which type?

kotykmax [81]

Answer:

B polynomial; binomial

Step-by-step explanation:

Yes, it is a polynomial, as it has two terms therefore it is a binomial.

4 0

3 years ago

Read 2 more answers

Test yourself 2

mafiozo [28]

Ok, so dy/dx=0 at the point (0,3) that is where x=0 and y=3.

$\int { 6x+6dx } \\ \\ =\frac { 6{ x }^{ 2 } }{ 2 } +6x+C\\ \\ =3{ x }^{ 2 }+6x+C$

$\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+C$

Now, f'(x)=0 when x=0.

Therefore:

$0=C\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x$

Now:

$\int { 3{ x }^{ 2 } } +6xdx\\ \\ =\frac { 3{ x }^{ 3 } }{ 3 } +\frac { 6x^{ 2 } }{ 2 } +C$

$={ x }^{ 3 }+3{ x }^{ 2 }+C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+C$

But when x=0, y=3, therefore:

$3=C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+3$

7 0

3 years ago

Read 2 more answers

The coordinates of point T are (0,4). The midpoint of Line ST is (2,-5). Find Coordinates of aS

Stels [109]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ S(\stackrel{x_1}{x}~,~\stackrel{y_1}{y})\qquad T(\stackrel{x_2}{0}~,~\stackrel{y_2}{4}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{0+x}{2}~~,~~\cfrac{4+y}{2} \right)~~=~~\stackrel{midpoint}{(2~,~-5)}\implies \begin{cases} \cfrac{0+x}{2}=2\\[1em] \boxed{x=4}\\ \cline{1-1} \cfrac{4+y}{2}=-5\\[1em] 4+y=-10\\ \boxed{y=-14} \end{cases}$

5 0

4 years ago