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Lina20 [59]
3 years ago
13

the bottom of a vase is a square.each side measury y+11 units.the square has a perimeter of 55 units what is the value of y?

Mathematics
2 answers:
Norma-Jean [14]3 years ago
8 0

Answer:

2.75

Step-by-step explanation:

55-11(4)=2.75

2.75+11=13.75

13.75x4=55

max2010maxim [7]3 years ago
6 0

Answer:

y = 11/4 or 2 3/4 units.

Step-by-step explanation:

Perimeter = 4 * length of 1 side

= 4(y + 11) , so

4(y + 11) = 55

4y + 44 = 55

4y = 11

y = 11/4 (answer).

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Find the area and perimeter of the following
suter [353]

Answer:

Perimeter is 8x²y+26x

Area is 52x³y

Step-by-step explanation:

The perimeter is 2W+2L and area is LW

Perimeter:

2(4x²y)+2(13x)

8x²y+26x

Area:

(4x²y)(13x)

52x³y

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2 years ago
Write 4.63 x 10^-3 in standard notation
Ierofanga [76]

Hello!

4.63 x 10^-3 = 0,00463

8 0
2 years ago
The graph shows f(x)and its transformation(x).
andre [41]

Using translation concepts, the equation for function g(x) given in the graph is:

g(x) = 2^{x + 2}

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

The parent function for this problem is given as follows:

f(x) = 2^x

From the graph, function g(x) was shifted 2 units left, hence x -> x + 2 and the definition is:

g(x) = 2^{x + 2}

More can be learned about translation concepts at brainly.com/question/28098112

#SPJ1

8 0
2 years ago
Q1:
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help
Gnesinka [82]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify :- 1 + - w² + 9w.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf1 + - w ^ { 2 } + 9 w

Quadratic polynomial can be factored using the transformation \sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \sf x_{1} and x_{2} are the solutions of the quadratic equation \sf \: ax^{2}+bx+c=0.

\large \sf-w^{2}+9w+1=0

All equations of the form \sf\:ax^{2}+bx+c=0 can be solved using the quadratic formula: \sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)}  \\

Square 9.

\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)}  \\

Multiply -4 times -1.

\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)}  \\

Add 81 to 4.

\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)}  \\

Multiply 2 times -1.

\large \sf \: w=\frac{-9±\sqrt{85}}{-2}  \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is plus. Add -9 to \sf\sqrt{85}.

\large \sf \: w=\frac{\sqrt{85}-9}{-2}  \\

Divide -9+ \sf\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is minus. Subtract \sf\sqrt{85} from -9.

\large \sf \: w=\frac{-\sqrt{85}-9}{-2}  \\

Divide \sf-9-\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}}  \\

Factor the original expression using \sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \sf\frac{9-\sqrt{85}}{2}for \sf\:x_{1} and \sf\frac{9+\sqrt{85}}{2} for \sf\:x_{2}.

\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

\large \sf \: 1 + -  w {}^{2}  + 9w \\  =\large  \boxed{\bf \: 1 -  {w}^{2}   + 9w}

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0
2 years ago
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