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hodyreva [135]
3 years ago
13

Karen spent 7/10 hour reading and 3/5 hours writing in her journal how much longer did karen spend reading than writing in her j

ournal
Mathematics
1 answer:
STatiana [176]3 years ago
3 0

Answer: 1/10

Step-by-step explanation:

From the question, we are informed that Karen spent 7/10 hour reading and 3/5 hours writing in her journal.

To calculate how much longer karen spent reading than writing in her journal, we have to subtract 3/5 from 7/10. This will be:

= 7/10 - 3/5

Note that the lowest common multiple of 5 and 10 is 10.

= 7/10 - 6/10

= 1/10

Therefore, the answer is 1/10

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Answer:

3-6=3

Step-by-step explanation:

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3 years ago
Which one is it? for a grade & due today!! help me please
Mama L [17]

Answer:the last one

Step-by-step explanation:

Divide 12 by 10.4

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It might come bigger out as a decimal 1.5

6 0
3 years ago
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please help :{ brainliest if you can get it right and I need you to prove why your answer is correct, thanksss​
beks73 [17]

Answer:

The Value of a=\frac{15}{4}.

Step-by-step explanation:

We have Named the figure please find the attachment for your reference.

Given:

PR = y

QR = a

RS = b

PS = z

PQ = x

QS = 15

∠P = 90°

∠R = 90°

∠Q = 60°

∠S = 30°

We need to find the Value of 'a'.

Solution:

Now we know that:

In Δ PQS

∠P = 90°

∠S = 30°

Now we know that;

sin\ \theta = \frac{opposite\ side}{Hypotenuse}

sin \ S= \frac{PQ}{QS}

Substituting the given values we get;

sin\ 30\°=\frac{x}{15}

Now we know that;

sin\ 30\° = \frac12

So we can say that;

\frac{1}{2}=\frac{x}{15}\\\\x=\frac{15}{2}

Now In Triangle PQR.

∠R = 90°

∠Q = 60°

So we can say that;

Cos \theta = \frac{adjacent \ Side}{Hypotenuse}\\

Cos\ Q = \frac{QR}{PQ}

Substituting the given values we get;

cos 60\°= \frac{a}{x}

Now we know that;

cos 60\°= \frac12

x=\frac{15}{2}

So substituting the values we get;

\frac{1}{2}=\frac{a}{\frac{15}{2}}

By Using Cross Multiplication we get;

a= \frac{1}{2}\times\frac{15}{2}\\\\a=\frac{15}{4}

Hence The Value of a=\frac{15}{4}.

6 0
3 years ago
If the function h is defined by h(x)=<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" a
harkovskaia [24]

Given:

The function is:

h(x)=x^2-3x+5

To find:

The value of h(2x+1).

Solution:

We have,

h(x)=x^2-3x+5

Putting x=2x+1, we get

h(2x+1)=(2x+1)^2-3(2x+1)+5

h(2x+1)=(2x)^2+2(2x)(1)+(1)^2-3(2x)-3(1)+5

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On combining like terms, we get

h(2x+1)=4x^2+(4x-6x)+(1-3+5)

h(2x+1)=4x^2-2x+3

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3 0
3 years ago
Please answer this question asap
lukranit [14]

Answer:

Use the formula

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I think it will help you.

8 0
2 years ago
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