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2 years ago

14

For a field trip, a high school wants to buy T-shirts and sweatshirts. The expression px+29y represents the cost of buying x T-s

hirts and y sweatshirts from a store. If each sweatshirt costs $17 more than a T-shirt, what is the value of p?

1 answer:

garri49 [273]2 years ago

7 0

Answer: UH

Step-by-step explanation:

You might be interested in

Q2) The isoscsles right triangle has 2 of it’s sides 10,10 it’s area is

nydimaria [60]

Answer:

Step-by-step explanation:

side a = 10

Area of isosceles triangle = $\frac{\sqrt{3}a^{2}}{4}\\\\$

= $\frac{\sqrt{3}*10*10}{4}\\\\=\sqrt{3}*5*5\\\\=25\sqrt{3}\\\\=25*1.732$

= 43.3 square units

3 0

2 years ago

Using a graphing utility, find the exact solutions of the system. Round to the nearest hundredth and choose a solution to the sy

Margaret [11]

Answer:

Part 1) The exact solutions are

$(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})$ and $(\frac{-1-\sqrt{21}} {2},4-\sqrt{21})$

Part 2) (1.79, 8.58)

Step-by-step explanation:

we have

$y=x^{2} +3x$ ----> equation A

$y=2x+5$ ----> equation B

we know that

When solving the system of equations by graphing, the solution of the system is the intersection points both graphs

<em>Find the exact solutions of the system</em>

equate equation A and equation B

$x^{2} +3x=2x+5\\x^{2} +3x-2x-5=0\\x^{2} +x-5=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2} +x-5=0$

so

$a=1\\b=1\\c=-5$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(-5)}} {2(1)}$

$x=\frac{-1\pm\sqrt{21}} {2}$

so

The solutions are

$x_1=\frac{-1+\sqrt{21}} {2}$

$x_2=\frac{-1-\sqrt{21}} {2}$

<em>Find the values of y</em>

<em>First solution</em>

For $x_1=\frac{-1+\sqrt{21}} {2}$

$y=2(\frac{-1+\sqrt{21}} {2})+5$

$y=-1+\sqrt{21}+5\\\\y=4+\sqrt{21}$

The first solution is the point $(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})$

<em>Second solution</em>

For $x_2=\frac{-1-\sqrt{21}} {2}$

$y=2(\frac{-1-\sqrt{21}} {2})+5$

$y=-1-\sqrt{21}+5\\\\y=4-\sqrt{21}$

The second solution is the point $(\frac{-1-\sqrt{21}} {2},4-\sqrt{21})$

Round to the nearest hundredth

<em>First solution </em>

$(\frac{-1+\sqrt{21}} {2},4+\sqrt{21})$ -----> $(1.79,8.58)$

$(\frac{-1-\sqrt{21}} {2},4-\sqrt{21})$ -----> $(-2.79,-0.58)$

see the attached figure to better understand the problem

6 0

3 years ago

Yalll n-4 = 3n + 6 do me a favah

Otrada [13]

<h2>Answer:</h2><h2>n=-5</h2>

n

-

4

=

3

+

6

n-4=3n+6

n−4=3n+6

Solve

1

Add

4

4

4

to both sides of the equation

−

4

=

3

+

6

n-4=3n+6

n−4=3n+6

−

4

+

4

=

3

+

6

+

4

n-4+{\color{#c92786}{4}}=3n+6+{\color{#c92786}{4}}

n−4+4=3n+6+4

2

Simplify

3

Subtract

3

3n

3n

from both sides of the equation

4

Simplify

5

Divide both sides of the equation by the same term

6

Simplify

5 0

2 years ago

Read 2 more answers

Please explain how to solve this (<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7Ba%5E5%20b%5E3%7D%20" id="TexFormula1" ti

olchik [2.2K]

2 is the base and ( $\sqrt[4][a^5 b^3]$ ) is the exponent.

the 4 is called an index and it means to fi d the fourth root of the expression under the ( $\sqrt[][text]) symbol.\\a fourth root is a factor that was multiplied four times. ex: 2*2*2*2=16. the fourth root of 16 is two. \\when you do square roots, you take the number representing the factor out of the radical. anything else that is not a square root stays under. ex: \\([tex]\sqrt[9][text])\\([tex]\sqrt[3*3*3][text]) there is one set of two 3's so\\([tex]3\sqrt[3][text])\\with a fourth root, you look for groups of four. the symbol of that group is placed outside the radical. anything else stays in.\\ ([tex]\sqrt[4][a^5 b^3]$ )

( $\sqrt[4][a*a*a*a*a*b*b*b]$ ) there is one group of 4 a's so

( $a\sqrt[4][a* b^3]$ )

there really isn't anything else to do to simplify the expression

5 0

2 years ago

storchak [24]

1. 7.00
2. A. <span>Sarah is paying the smaller price of $1.25 per cupcake.

</span>

6 0

2 years ago