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2 years ago

Partial products of 310x434

Mathematics

2 answers:

Vanyuwa [196]2 years ago

6 0

It's 134,540 and use photomath app it helps a lot

Rainbow [258]2 years ago

6 0

The answer is 134,540

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If line AD is a tangent to circle B at point C, and m∠ABC = 55º, what is the measure of ∠BAD?

4vir4ik [10]

The tangent AD at point C forms a right angle to circle B, such as ∠ACB=90° and ∠ABC=55°.

We also know that the summation of all angles in a triangle is 180°. Thus we have the expression below:
180°=∠ABC + ∠ACB + ∠BAD
180°=55°+90°+∠BAD
∠BAD=35°

Therefore, the answer is 35°.

7 0

3 years ago

Read 2 more answers

Can you answer this quick it's so hard

Bingel [31]

Answer:

square

triangle

square pyramid

find area of base and area of a lateral face (base is 1 side squared, lateral face is base * height / 2. Multiply area of lateral face by 4 and add the product to the area of the base to find the Surface Area.

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Roberta Arthur had the following reciepts for the week of April 12th: 50$ gift; 25$ rebate; April 13th, 2.45$ refund on aluminum

oksian1 [2.3K]

Answer:

$329.44

Step-by-step explanation:

$50-25= 25

25.00-2.45=22.55

299.89+22.55=322.44

322.44+7.00= 329.44

I didnt see april 15 if its supposed to be added i there just add 299.89 to the total.

7 0

3 years ago

Anyone can help me with this

Leokris [45]

Given:

The data values are

11, 12, 10, 7, 9, 18

To find:

The median, lowest value, greatest value, lower quartile, upper quartile, interquartile range.

Solution:

We have,

11, 12, 10, 7, 9, 18

Arrange the data values in ascending order.

7, 9, 10, 11, 12, 18

Divide the data in two equal parts.

(7, 9, 10), (11, 12, 18)

Divide each parenthesis in 2 equal parts.

(7), 9, (10), (11), 12, (18)

Now,

Median = $\dfrac{10+11}{2}$

= $\dfrac{21}{2}$

= $10.5$

Lowest value = 7

Greatest value = 18

Lower quartile = 9

Upper quartile = 12

Interquartile range (IQR) = Upper quartile - Lower quartile

= 12 - 9

= 3

Therefore, median is 10.5, lowest value is 7, greatest value is 18, lower quartile 9, upper quartile 12 and interquartile range is 3.

8 0

3 years ago

Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol

koban [17]

Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

4 0

2 years ago