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rodikova [14]
3 years ago
11

which set of side lengths could not be used form atriangle? a. 5cm, 7 cm , 10 cm b. 4 cm , 10cm, 15cm c. 3cm, 8 cm , 6cm d. 12cm

, 8cm , 9 cm
Mathematics
1 answer:
emmasim [6.3K]3 years ago
8 0
B. 4 cm, 10 cm, and 15 cm

According to two sides, 4, and 10 cm, the third side has to be between 
10-4 and 10+4
6<side length<14

However, the side length is 15 cm, greater than 14 cm, so it is not a triangle.

Hope this helps :)
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.75

Step-by-step explanation:

because if you turn .75 into a fraction it is 1/4

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Rolling two die, how many ways can a five show up on at least one die
IRINA_888 [86]
Total possibilities would be: 

(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

You want 5 on at-least one, so mark 5th column & 5th raw.
There are 6 in both lines with 1 common = (5,5)

In short, Your Answer would be 11

Hope this helps!
3 0
3 years ago
Write the expression using rational exponents. Then simplify and convert back to radical notation.
ioda

Answer:

The radical notation is 3x\sqrt[3]{y^2z}

Step-by-step explanation:

Given

\sqrt[3]{27 x^{3} y^{2} z}

Step 1 of 1

Write the expression using rational exponents.

\sqrt[n]{a^{m}}=\left(a^{m}\right)^{\frac{1}{n}}

=a^{\frac{m}{n}}:\left({27 x^{3} y^{2} z})^{\frac{1}{3}}

$(a \cdot b)^{r}=a^{r} \cdot b^{r}:(27)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

=$(3^3)^{\frac{1}{3}}\left(x^{3}\right)^{\frac{1}{3}} \cdot\left(y^{2}\right)^{\frac{1}{3}} \cdot(z)^{\frac{1}{3}}$

$=\left(3\right)\left(x}\right)} \cdot\left(y}\right)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

Simplify $3 x \cdot(y)^{\frac{2}{3}} \cdot(z)^{\frac{1}{3}}$

$=3 x \sqrt[3]{y^{2} z}$

Learn more about radical notation, refer :

brainly.com/question/15678734

4 0
3 years ago
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