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3 years ago

You are asked to draw a triangle with side lengths of 3 inches, 4 inches, and 1 inch. How many triangles like this can you draw?

1 answer:

6 0

Given the sides 3 inches, 4 inches and 1 inch, the possible number of triangle that you can draw is only 1. Since a triangle has 3 sides, then you are given 3 sides, then the possibility is 1. To check whether the sides are legitimate to form a triangle, following a theorem, the sum of two sides must be greater than the third side.

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Is .8749 greater than or less than .05

Leto [7]

Answer:

it is greater than .05 because 8 is larger

6 0

3 years ago

Pi High school ordered 40 science books. The next week, the school ordered 30 algebra books. The bill for the first order was $3

Darina [25.2K]

40s=360+30a
40s+30a=$3960

40s=360+30a
40s/40=360+30a/40
s = 9+3/4a

40(9+3/4a) +30a=$3960
360+30a+30a=$3960
360+60a=$3960
360-360+60a=$3960
60a=$3960
60a/60 = $3960/60
a= $60
The algebra book is $60

4 0

3 years ago

Please help me with Algebra 2 questions (2).

user100 [1]

8. The simplified form of the expression is $- 5 x^{\frac{4}{5}}y^{3}$

11. The value of x is 4

<h3>Simplifying an expression</h3>

From the question, we are to simplify the give expression

The given expression is

$y^{3}\sqrt[5]{32x^{4} } - 7\sqrt[5]{x^{4}y^{15} }$

$y^{3} (32x^{4} )^{\frac{1}{5} } - 7(x^{4}y^{15} )^{\frac{1}{5}$

$y^{3} \times 32^{\frac{1}{5} } x^{\frac{4}{5} } - 7 \times x^{\frac{4}{5}}y^{\frac{15}{5}}$

$y^{3} \times 2^{5 \times} ^{\frac{1}{5} } \times x^{\frac{4}{5} } - 7 \times x^{\frac{4}{5}}y^{3}$

$y^{3} \times 2 \times x^{\frac{4}{5} } - 7 \times x^{\frac{4}{5}}y^{3}$

$2 x^{\frac{4}{5} } y^{3} - 7 x^{\frac{4}{5}}y^{3}$

$= - 5 x^{\frac{4}{5}}y^{3}$

11. $\frac{1}{2}(5x+7 )^{\frac{2}{3} } =\frac{9}{2}$

Multiply both sides by 2

$2 \times \frac{1}{2}(5x+7 )^{\frac{2}{3} } =2 \times \frac{9}{2}$

$(5x+7 )^{\frac{2}{3} } =9$

Multiply both sides by a power of 3/2

$(5x+7 )^{\frac{2}{3} \times {\frac{3}{2} } } =9^{\frac{3}{2} }$

$(5x+7 )=3^{2 \times} ^{\frac{3}{2} }$

$5x+7 =3^{3}$

$5x+7 =27$

$5x =27-7$

$5x = 20$

$x = \frac{20}{5}$

x = 4

Hence,

8. The simplified form of the expression is $- 5 x^{\frac{4}{5}}y^{3}$

11. The value of x is 4

Learn more on Simplifying expressions here: brainly.com/question/11688560

#SPJ1

4 0

2 years ago

Myranda received a movie gift card for $100 to her

PSYCHO15rus [73]

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

x ---> represents the number of matinee tickets she could purchase

y ---> represents the number of evening tickets she could purchase

we know that

The number of matinee tickets (x) multiplied by $7.50 each plus the number of evening tickets (y) multiplied by $12.50 each must be at most (less than or equal to) $100

The inequality that represent this problem is

$7.50x+12.50y\leq 100$

using a graphing tool

The solution is the triangular shaded area

Remember that the number of tickets must be a whole number

see the attached figure

5 0

3 years ago

Johan's gift shop needs to mail 45 checks to the bank. If they can put 5 checks in each envelope, how many envelopes how does th

Iteru [2.4K]

Answer: 9 envelopes

Step-by-step explanation:

Divide the 45 checks into 9 groups of 5.

------------------------------------------------------------------------

| 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 | 5 |

------------------------------------------------------------------------

45 checks divided into 9 envelopes equals 5 checks per envelope

45 ÷ 5 = 9

Take away one group of 5 checks from the 45 checks. Then take away another group of 5 checks, and another, and another. When there are no more checks, count how many groups you took away. You can take 9 groups away from 45, so that is your answer.

To tell that the answer is 9 groups of 5 checks, follow these steps.

Look at the equation. 45 ÷ 5 = 9.

The amount of groups is what 45 ÷ 5 equals. What the 45 checks is divided by is the amount in each group.

This means that there are 9 groups of 5 checks.

8 0

3 years ago