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Amanda [17]
2 years ago
13

Which of these is a correct statement?

Mathematics
1 answer:
Allushta [10]2 years ago
6 0
I got work on my B I got a A
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Can someone help me with 4-6 plz
satela [25.4K]

Step-by-step explanation:

4. To use SSS, you need three pairs of congruent sides.  You're given two pairs of congruent sides, so the additional information needed is WY ≅ KM.

5. To use ASA, you need a pair of congruent sides between two pairs of congruent angles.  You're given one pair of congruent angles, and since the triangles share a common side, we know BC ≅ BC.  So the additional information needed is ∠WBC ≅ ∠ACB.

6. To use SAS, you need a pair of congruent angles between two pairs of congruent sides.  You're given two pairs of congruent sides, so the additional information needed is ∠I ≅ ∠F.

4 0
3 years ago
In a positive correlation, __________. A. the independent variable is always negative B. an increase in the independent variable
lyudmila [28]
The answer would be, "B".
8 0
3 years ago
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Round 8,503 to nearest thousand
Jlenok [28]

Answer:

9,000

Step-by-step explanation:

If x < 5, we round down.

If x ≥ 5, we round up.

Since 8,503, 500 is ≥ 5, we round up to 9. So,

8,503 ≈ 9,000

4 0
3 years ago
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Find the value of x in this figure.
AleksAgata [21]

Step-by-step explanation:

angle x is equal to 65

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3 0
3 years ago
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At the beginning of year 1, Sam invests $700 at an annual compound interest
Aleksandr-060686 [28]

at the beginning of year 4, only 3 years have elapsed, the 4th year hasn't started yet, since it's at the beginning, so at the beginning of year 4 we can say only 4-1 years have elapsed.

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$700\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=\textit{elapsed years}\dotfill &3 \end{cases}

A=700\left( 1 + \frac{0.05}{1} \right)^{1\cdot 3}\implies A = 700(1+0.05)^3\implies A(4)=700(1+0.05)^{4-1} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill A(n)=700(1+0.05)^{n-1}~\hfill

6 0
2 years ago
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