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Mars2501 [29]
4 years ago
12

5/-7x (-9y/8) multiply and simplify ​

Mathematics
1 answer:
LekaFEV [45]4 years ago
7 0

Answer:

\large\boxed{\dfrac{5}{-7x}\cdot\dfrac{-9y}{8}=\dfrac{45y}{56x}}

Step-by-step explanation:

\dfrac{5}{-7x}\cdot\dfrac{-9y}{8}=\dfrac{(5)(-9y)}{(-7x)(8)}=\dfrac{-45y}{-56x}=\dfrac{45y}{56x}

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Plz help me plz real quick please​
STatiana [176]
The answer out of a and b
7 0
3 years ago
Holly had 5,000 in her bank account . She withdrew $800 to buy a new bike . What is the percent decrease in the blance of her ac
ser-zykov [4K]
$800 is 16 percent of $5000, so the percentage decrease in the balance of her bank account is 16 percent.
5 0
4 years ago
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the two numbers.
Naddik [55]
Sol. (1) Prime factors of 26 = 2 x 13
Prime factors of 91 = 7 x 13
Hence, HCF = Common factors between 26 and 91 =13 and LCM=13x2x7=182

Now product of numbers 26 and 91
= 26 x 91 = 2366 and Product of HCF and LCM = 13 x 182 = 2366

So, it verify that product of two numbers = Product of HCF and LCM.

(2) Prime factors of 510 = 2 x 3 x 5 x 17
Prime factors of 92 = 2 x 2 x 23
Hence,HCF=2 and LCM=2×2×3 × 5×17×23 =23460

Now product of Numbers 510 and 92 = 46920 and product of HCF and LCM = 2 x 23460= 46920

Hence, verified that product of two numbers 18 equal to product of their HCF and LCM.

(3) Prime factors of336 = 2 x 2 x 2 x 2 x 3 x 7

Prime factors of 54 = 2 x 3 x 3 x 3
Hence, HCF (Product of common factors of 336 and 54)
=2 x 3=6

And LCM (Product of all common factors with remaining factors)
=(2 x 3)x 2 x 2 x 2 x 3 x 3 x 7=3024

Now, product of numbers 336 and 54 = 336 x 54 = 18144
and product of HCF and LCM = 6 x 3024 = 18144
Hence, product of two numbers: Product of HCF and LCM.
5 0
3 years ago
Simplify.<br>2x – 15 + 3x – 33 - 10​
solong [7]

Answer:

5x - 58

Step-by-step explanation:

6 0
4 years ago
Read 2 more answers
Find the diagonal of a cube if its side equals 5. When applicable, simplify radicals and show all of your work. Help plz
9966 [12]
1. Hi, please take a good look at the pictures at the bottom.

2. What we are trying to find here is BH. 

3. The second picture shows the "diagonal section", DBFH in another perspective. 

4. We notice that HB is the hypothenuse of right triangle DBH.

5. So we can apply the pythegoran theorem to find HB, but first we need to find HD and DB.

6. HD is just a side of the cube so it is 5 units. DB is the diagonal of the base square  ABCD, so we find it by applying the pythagorean theorem in right triangle ABD: 
AB^2+AD^2=BD^2

5^2+5^2=BD^2&#10;&#10;2*5^2=BD^2&#10;&#10; \sqrt{2*5^2} = \sqrt{BD^2} &#10;&#10;BD=5 \sqrt{2} &#10;

6. So we have both HD and DB:

HD^2+DB^2=HB^2&#10;&#10;5^2+ (5 \sqrt{2} )^{2}=HB^2&#10;&#10;25+50=HB^2&#10;&#10;HB^2=75&#10;&#10;HB= \sqrt{25*3}=    \sqrt{25} * \sqrt{3}= 5\sqrt{3}

7. Remark: Another way would be to use the generalized Pythagorean theorem, which is as follows:

Diagonal^2=5^2+5^2+5^2&#10;&#10;Diagonal =  \sqrt{3* 5^{2} } =5 \sqrt{3}


8. In general, given a right rectangular prism with sides a, b and c, the diagonal is given by \sqrt{a^2+b^2+c^2}

3 0
3 years ago
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