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3 years ago

The estimated sum of 857 + 234 is

Mathematics

1 answer:

Llana [10]3 years ago

6 0

Answer:

1,091

hope it helps.

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Selena was comparing the flying speeds of different birds in science class. Which of the three birds flies fastest? Why?

Mashcka [7]

Answer:

Option C. The robin flies fastest because its rate is 15 m/s

Step-by-step explanation:

Let

x -----> the time in seconds

y ----> the distance in meters

The speed is equal to divide the distance by the time

In this problem, the slope or unit rate of the linear equation is the same that the speed

<em>Robin's Flight</em>

<em>Find the slope</em>

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

take two points from the table

(2,30) and (3,45)

substitute in the formula

$m=\frac{45-30}{3-2}$

$m=\frac{15}{1}$

$m=15\frac{m}{sec}$

<em>Cardinal's Flight</em>

$y=10x$

$m=10\frac{m}{sec}$

<em>Blue Jay's Flight</em>

take two points from the graph

(0,0) and (2,20)

substitute in the formula

$m=\frac{20-0}{2-0}$

$m=\frac{20}{2}$

$m=10\frac{m}{sec}$

Compare

The robin flies fastest because its rate is 15 m/s

4 0

3 years ago

Mr. Abaya goy P700,000 loan for the expansion of his business payable monthly in 4years. How much os the monthly amortization if

Vaselesa [24]

The monthly amortization is P23,511.63

<h2>Compound interest</h2>

The formula for calculating the compound interest is expressed as:

$A = P(1+r/n)^t$

whrere:

P is the principal = P700,000
r is the rate = 12% = 0.12
t is the time = 4 years
n = 12

Substitute the parameters

$A = 700000(1+0.12/12)^{4(12)}\\
A = 700,000(1.01)^{48}\\
A =1,128,558.25$

Calculate the monthly payment:

Monthly payment = $\frac{1,128,558.25}{48} =23,511.63$

Hence the monthly amortization is P23,511.63

Learn more on compound interest here: brainly.com/question/24924853

8 0

2 years ago

1. L.B. Johnson Middle School held a track and field event during the school year. The chess club sold various drink and snack i

iris [78.8K]

Divide $2,673 by 486

6 0

3 years ago

Read 2 more answers

Help please any One who can solve this equation? ☺

malfutka [58]

If y varies directly with x, that means they change at proportion rates. We can define their relationship using a constant k.

y = kx

Plug in what we know to find k.

-3 = 5k

k = -3/5

So:

y = -1 * (-3/5)

y = 3/5

6 0

3 years ago

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