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2 years ago

The radius of the large sphere is double the radius of the small sphere.

Mathematics

1 answer:

kkurt [141]2 years ago

4 0

Step-by-step explanation:

the answer is in the image above

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Lara created a 15-pound platter of three different types of cookies for a banquet. The peanut butter cookies sell for $2 per pou

Bond [772]

Answer:

peanut butter cookies:8

sugar cookies:2

chocolate chip cookies:5

ratio: 4:1:2.5

Step-by-step explanation:

x=peanut butter cookies

y=sugar cookies

a=chocolate chip cookies

40=2x+3y+3.6a

15=x+y+a

x=a+3

15=(a+3)+y+a

y=12-2a

40=2(a+3)+3(12-2a)+3.6a

a=5

x=8

y=2

5 0

3 years ago

Whats 4 2/5 x 1 1/2 pls help

Veronika [31]

The answer is that (It’s 6.6)

8 0

2 years ago

Read 2 more answers

Hey can you please help me posted picture of question

marta [7]

Two transformations occur on F(x) to obtain G(x).

a) Multiplication by -1
This means, the function is reflected across x-axis

b) Multiplication by 5.
This means vertical stretch by a factor of 5.

So, G(x) can be obtained by reflecting F(x) across x-axis and stretching it vertically be a factor of 5.

So, the correct answer is option B

7 0

2 years ago

Read 2 more answers

Prove A-(BnC) = (A-B)U(A-C), explain with an example

NikAS [45]

Answer:

Prove set equality by showing that for any element $x$ , $x \in (A \backslash (B \cap C))$ if and only if $x \in ((A \backslash B) \cup (A \backslash C))$ .

Example:

$A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace$ .

$B = \lbrace0,\, 1 \rbrace$ .

$C = \lbrace0,\, 2 \rbrace$ .

$\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}$ .

$\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}$ .

Step-by-step explanation:

Proof for $[x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))]$ for any element $x$ :

Assume that $x \in (A \backslash (B \cap C))$ . Thus, $x \in A$ and $x \not \in (B \cap C)$ .

Since $x \not \in (B \cap C)$ , either $x \not \in B$ or $x \not \in C$ (or both.)

If $x \not \in B$ , then combined with $x \in A$ , $x \in (A \backslash B)$ .
Similarly, if $x \not \in C$ , then combined with $x \in A$ , $x \in (A \backslash C)$ .

Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ as required.

Proof for $[x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]$ :

Assume that $x \in ((A \backslash B) \cup (A \backslash C))$ . Thus, either $x \in (A \backslash B)$ or $x \in (A \backslash C)$ (or both.)

If $x \in (A \backslash B)$ , then $x \in A$ and $x \not \in B$ . Notice that $(x \not \in B) \implies (x \not \in (B \cap C))$ since the contrapositive of that statement, $(x \in (B \cap C)) \implies (x \in B)$ , is true. Therefore, $x \not \in (B \cap C)$ and thus $x \in A \backslash (B \cap C)$ .
Otherwise, if $x \in A \backslash C$ , then $x \in A$ and $x \not \in C$ . Similarly, $x \not \in C \!$ implies $x \not \in (B \cap C)$ . Therefore, $x \in A \backslash (B \cap C)$ .

Either way, $x \in A \backslash (B \cap C)$ .

Therefore, $x \in ((A \backslash B) \cup (A \backslash C))$ implies $x \in A \backslash (B \cap C)$ , as required.

8 0

2 years ago

Please help me with B

QveST [7]

She is incorrect because 27 is not a prime number but yet it has 7 at the end

8 0

2 years ago