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3 years ago

10

40/20 x 1/6 does it equal 40/120 im doing another problem and this is a step just checking my answer

1 answer:

Hitman42 [59]3 years ago

4 0

yeah it does...

..........................

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Use the net to find the lateral area of the cylinder.

ioda

The lateral area is the area around, the bases are the circles so the lateral area is the area of the rectangle

we just need the legnth of the rectangle which is the circumfernece of the circle

C=pid
d=12
C=12pi

area=lw
aera=12pi*15
area=180pi yd^2

4 0

4 years ago

Read 2 more answers

1. The length and width of a certain rectangle with an area of 12 units² can be expressed as x + 2 and

Katen [24]

$~~~~(x+2)(x+3) = 12\\\\\implies x^2+3x+2x+6 -12=0\\ \\\implies x^2 +5x -6 = 0\\\\\implies x^2 +6x-x-6 = 0\\\\\implies x(x+6) -(x+6) =0\\\\\implies (x+6)(x-1)=0\\\\\implies x=1~~~~~~~~~;[x\neq -6~ \text{because length cannot be negative.}}]$

3 0

3 years ago

Determine the value of x and y

lapo4ka [179]

Step-by-step explanation:

IGH = JKL

GH = KL

12 = 36

12 × x = 36

x = 36 ÷ 12 = 3

IG =

$\sqrt{ {16 {}^{2} + 12}^{2} }$

= 20

y = 16 × 3 = 48

x =

$\sqrt{ {48}^{2} + {36}^{2} }$

= 60

6 0

3 years ago

Law of radicals

andrey2020 [161]

Answer:

1) $\sqrt{x^7}=x^{\frac{7}{2}$

2) $\sqrt[3]{y^5}=y^{\frac{5}{3}$

3) $\sqrt[5]{a^{12}}=a^{\frac{12}{5} }$

4) $\sqrt[4]{z^{9}}=z^\frac{9}{4}$

Step-by-step explanation:

1) $\sqrt{x^7}$

We know that $\sqrt{x}=x^{\frac{1}{2}$

So, $\sqrt{x^7}=x^{\frac{7}{2}$

2) $\sqrt[3]{y^5}$

We know that $\sqrt[3]{x}=x^{\frac{1}{3}$

So, $\sqrt[3]{y^5}=y^{\frac{5}{3}$

3) $\sqrt[5]{a^{12}}$

We know that $\sqrt[5]{x}=x^{\frac{1}{5}$

So, $\sqrt[5]{a^{12}}=a^{\frac{12}{5} }$

4) $\sqrt[4]{z^{9}}$

We know that $\sqrt[4]{x}=x^{\frac{1}{4}$

So, $\sqrt[4]{z^{9}}=z^\frac{9}{4}$

6 0

3 years ago

From a bowl containing five red, three white, and seven blue chips, select four at random and without replacement. Compute the c

snow_lady [41]

Answer:

The probability is $\frac{5}{9}$

Step-by-step explanation:

Let A be the event of one red, zero white, and three blue chips,

And, B is the event of at least three blue chips,

Since, A ∩ B = A (because If A happens that it is obvious that B will happen )

Thus, the conditional probability of A if B is given,

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$

Now, red chips = 5,

White chips = 3,

Blue chips = 7,

Total chips = 5 + 3 + 7 = 15

Since, the probability of one red, zero white, and three blue chips, when four chips are chosen,

$P(A)=\frac{^5C_1\times ^3C_0\times ^7C_3}{^{15}C_4}$

$=\frac{5\times 35}{1365}$

$=\frac{175}{1365}$

$=\frac{5}{39}$

While, the probability that of at least three blue chips,

$P(B)=\frac{^8C_1\times ^7C_3+^8C_0\times ^7C_4}{^{15}C_4}$

$=\frac{8\times 35+35}{1365}$

$=\frac{315}{1365}$

$=\frac{3}{13}$

Hence, the required conditional probability would be,

$P(\frac{A}{B})=\frac{5/39}{3/13}$

$=\frac{65}{117}$

$=\frac{5}{9}$

4 0

3 years ago