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3 years ago

Draw a Venn diagram of the Gcf of 154 and 210

Mathematics

1 answer:

ruslelena [56]3 years ago

5 0

See attachment

each circle is the factor

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A bricklayer lays 421 bricks on day 1 of a project, 623 on the 2nd day, and 343 the 3rd day, how many was the total number of br

Angelina_Jolie [31]

1,387 because you just add all the bricks (421+623+343)

4 0

3 years ago

Use this picture to solve for X

Zepler [3.9K]

Answer:

x=2

Step-by-step explanation:

i would set up a system of equation using the pythagorean theorem:

1^2+x^2=z^2

4^2+x^2=y^2

Add up these equations to get:

17+2x^2=y^2+z^2

But notice that

y^2+z^2=5^2

So you can substitute

17+2x^2=25

2x^2=8

x^2=4

x=2

8 0

2 years ago

On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)

xz_007 [3.2K]

Answer:

$P = 10 + 2\sqrt{53}$ units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3), X (2, 3),

Y (4, -4), Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

For WX:

$(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)$

$WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}$

$WX = \sqrt{(-5)^2 + (0)^2}$

$WX = \sqrt{25}$

$WX = 5$

For XY:

$(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)$

$XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}$

$XY = \sqrt{-2^2 + (3 +4)^2}$

$XY = \sqrt{-2^2 + 7^2}$

$XY = \sqrt{4 + 49}$

$XY = \sqrt{53}$

For YZ:

$(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)$

$YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}$

$YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}$

$YZ = \sqrt{7^2 + (-2)^2}$

$YZ = \sqrt{49 + 4}$

$YZ = \sqrt{53}$

For ZW:

$(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)$

$ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}$

$ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}$

$ZW = \sqrt{0^2 + (-5)^2}$

$ZW = \sqrt{0 + 25}$

$ZW = \sqrt{25}$

$ZW = 5$

The Perimeter (P) is as follows:

$P = WX + XY + YZ + ZW$

$P = 5 + \sqrt{53} + \sqrt{53} + 5$

$P = 5 + 5 + \sqrt{53} + \sqrt{53}$

$P = 10 + 2\sqrt{53}$ units

6 0

3 years ago

Could you please help with this question for domain, range, and end behavior? I started it, but I think I'm doing it wrong.

Tanzania [10]

It looks like you have the domain confused for the range! You can think of the domain as the set of all "inputs" for a function (all of the x values which are allowed). In the given function, we have no explicit restrictions on the domain, and no situations like division by 0 or taking the square root of a negative number that would otherwise put limits on it, so our domain would simply be the set of all real numbers, R. Inequality notation doesn't really use ∞, so you could just put an R to represent the set. In set notation, we'd write

$\{x\in\re\mathbb{R}\}$

and in interval notation,

$(-\infty,\infty)$

The <em>range</em>, on the other hand, is the set of all possible <em>outputs</em> of a function - here, it's the set of all values f(x) can be. In the case of quadratic equations (equations with an x² term), there will always be some minimum or maximum value limiting the range. Here, we see on the graph that the maximum value for f(x) is 3. The range of the function then includes all values less than or equal to 3. As in inequality, we can say that

$f(x)\leq3$ ,

in set notation:

$\{f(x)\in\mathbb{R}\ |\ f(x)\leq3\}$

(this just means "f(x) is a real number less than or equal to 3")

and in interval notation:

$(-\infty,3]$

8 0

3 years ago

Find the difference using suitable identity (10.3 × 9.7) – (10.7 × 9.3)

finlep [7]

Answer:

$\small\textrm\red{-------------------------------------------------------------------}$ $\small\textrm\blue{-------------------------------------------------------------------}$ $\small\textrm\green{-------------------------------------------------------------------}$ $\small\textrm\blue{-------------------------------------------------------------------}$ mag sayaw ka nalang

$\small\textrm\green{-------------------------------------------------------------------}$ $\small\textrm\blue{-------------------------------------------------------------------}$ $\small\textrm\blue{-------------------------------------------------------------------}$ $\small\textrm\red{-------------------------------------------------------------------}$

5 0

3 years ago