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Alika [10]
2 years ago
9

If Ronaldo scored 78 out of 90 in his test, what percentage did he get?​

Mathematics
2 answers:
denpristay [2]2 years ago
6 0

Answer:

86.67%

Step-by-step explanation:

\frac{100\%}{x\%}=\frac{90}{78}

Take the inverse/reciprocal:

\frac{x\%}{100\%}=\frac{78}{90}

x=86.67%

Hope this helps!

nignag [31]2 years ago
6 0

Answer:

86.67%

Step-by-step explanation:

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Lois and Joe start saving pennies on new years day. Lois has 5 pennies and saves 2 pennies every day. Joe starts with 4 pennies
Mazyrski [523]
Well Your answer is 32 beacuse if lois has 5 pennies and saves 2 and joe start with 4 pennies and saves 6. well what i do first is 2 over 5 and then i do 4 over 6 and then i multiply 5 and 6 and then it 30 and that goes on the bottom and then i multiply 2×6 =12 and then i multiply 4×5=20 so i add both answer and it get me 30 that your answer
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3 years ago
What is the effect on the graph of y=-2x+7 if the slope changes to 2
Tatiana [17]
In y=-2x+7 the graph will rise from right to left while in y=2x+7 the graph will rise from left to right. Which gives us in conclusion of the slope is negative the graph line will rise from right to left and if positive then left to right
3 0
2 years ago
√48a^4 b^5 c^3 d simplify and keep it in radical form
Morgarella [4.7K]
I will assume the square root extends all the way across.

\sqrt{48a^4b^5c^3}

\sqrt{2^4\cdot3\cdot a^4b^5c^3}

\sqrt{(2^2)^2\cdot3\cdot(a^2)^2\cdot(b^2)^2\cdot b\cdot c^2\cdot c}

\sqrt{(4)^2\cdot3\cdot(a^2)^2\cdot(b^2)^2\cdot b\cdot c^2\cdot c}

\sqrt{(4\cdot a^2 \cdot b^2 \cdot c)^2\cdot3\cdot b\cdot c}

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\boxed{4a^2b^2c\sqrt{3bc}}






3 0
3 years ago
Fred bought nine new collectable cards to add to his collection. The next day his dog ate half of his collection.
Anna [14]

Answer:

half

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

6 0
2 years ago
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