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3 years ago

6

Solve for x. Round to the nearest tenth of a degree, if necessary.

1 answer:

devlian [24]3 years ago

5 0

Answer:

59

Step-by-step explanation:

do the adding up of all angles you will get a number greater than 59 and divide the number withh the amountt of angles there are which are 3 so divide the number by three and you will end up with 59 and thats the answer

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I have over 20 pencils in my desk

AleksAgata [21]

Answer:

that means you probably have 40 or 30

Step-by-step explanation:

6 0

3 years ago

Choose the list that shows these numbers in order from smallest to largest.

svetlana [45]

Answer:

0.3, 0.32, 0.321, 0.345

Step-by-step explanation:

The maximum number of decimals shown in the series is 3.

Therefore, let's rewrite all the numbers with three decimal places.

0.32, 0.321, 0.3, 0.345 can be written as

0.320, 0.321, 0.300, 0.345

Now, we can easily compare the numbers and we can conclude:

0.300 < 0.320 < 0.321 < 0.345

Hence, the smallest to largest order is:

0.3, 0.32, 0.321, 0.345

4 0

3 years ago

Graph the system of equations, and determine the solution.<br> x+4y=8

bekas [8.4K]

Answer:

12

Step-by-step explanation:

5 0

3 years ago

Which statement is true about the result of a rigid transformation?

Ulleksa [173]

I think the answer is a

6 0

3 years ago

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

3 years ago