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2 years ago

PLZZZ HELPPPP ILL MARK BRAINLIEST TO FIRST AND CORRECT ANSWER THAT HAS WORK AND A THX TO SECOND AND CORRECT ANSWER.

Mathematics

1 answer:

vovikov84 [41]2 years ago

6 0

Answer: m = 20.

Step-by-step explanation: A triangle has to be 180 degrees. 180-135-25=20.

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Please help me with this Algebra problem: The length of a rectangle is 91 less than six times the width of the rectangle. If the

horsena [70]

182 Let x = width then y = 6x - 91. P (perimeter) = 2x + 2y = 84 then

2(x) + 2(6x - 91) = 84 | 2x + 12x - 182 =84 | 14x = 84 + 182 | 14x = 266 | x = 19 cm

y = 6x - 91 | y = 6 (19) - 91 | y = 114 - 91 | y = 23 cm

Check: P = 2x + 2y | P = 2(19) + 2(23) | P = 38 + 46 | P = 84 cm (Checked)

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3 years ago

If a/ b = 6/7 and a/c = 2/5 , what is the value of 3b+c in terms of a?

Nookie1986 [14]

Not sure. has the same question

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3 years ago

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In ATUV, the measure of V=90°, TU = 5.4 feet, and VT = 1.6 feet. Find the measure

luda_lava [24]

Answer:73

Step-by-step explanation:

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3 years ago

DIscrete Math

Daniel [21]

Answer:

Step-by-step explanation:

As the statement is ‘‘if and only if’’ we need to prove two implications

$f : X \rightarrow Y$ is surjective implies there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ .
If there exists a function $h : Y \rightarrow X$ such that $f\circ h = 1_Y$ , then $f : X \rightarrow Y$ is surjective

Let us start by the first implication.

Our hypothesis is that the function $f : X \rightarrow Y$ is surjective. From this we know that for every $y\in Y$ there exist, at least, one $x\in X$ such that $y=f(x)$ .

Now, define the sets $X_y = \{x\in X: y=f(x)\}$ . Notice that the set $X_y$ is the pre-image of the element $y$ . Also, from the fact that $f$ is a function we deduce that $X_{y_1}\cap X_{y_2}=\emptyset$ , and because $f$ the sets $X_y$ are no empty.

From each set $X_y$ choose only one element $x_y$ , and notice that $f(x_y)=y$ .

So, we can define the function $h:Y\rightarrow X$ as $h(y)=x_y$ . It is no difficult to conclude that $f\circ h(y) = f(x_y)=y$ . With this we have that $f\circ h=1_Y$ , and the prove is complete.

Now, let us prove the second implication.

We have that there exists a function $h:Y\rightarrow X$ such that $f\circ h=1_Y$ .

Take an element $y\in Y$ , then $f\circ h(y)=y$ . Now, write $x=h(y)$ and notice that $x\in X$ . Also, with this we have that $f(x)=y$ .

So, for every element $y\in Y$ we have found that an element $x\in X$ (recall that $x=h(y)$ ) such that $y=f(x)$ , which is equivalent to the fact that $f$ is surjective. Therefore, the prove is complete.

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2 years ago

Express the product of 2x^2+6-8 and x+3 in standered form ASap plz !!!!

katrin [286]

1095 to the second power is what google told me!

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3 years ago