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3 years ago

Sandy had 219 football cards. She bought 57 more at a garage sale. How many did she have then?

Mathematics

2 answers:

Blizzard [7]3 years ago

8 0

Answer:

276

Step-by-step explanation:

add 219 and 57 together

8_murik_8 [283]3 years ago

5 0

Therefore, Sandy has 276 football cards.

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Komok [63]

Answer:

y=x-3/4

Step-by-step explanation:

y-int is b (y=mx+b) so the slope is m. all u have to do is plug in

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2 years ago

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A total of $37185 is invested at an annual interest rate of 2% compounded quarterly. Find the balance after 11 years.

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A=37,185×(1+0.02÷4)^(4×11)
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2 years ago

Question attached plz answer

Bezzdna [24]

Answer:

x = - 2.5

Step-by-step explanation:

Given that the sketch represents

y = x² + bx + c

The graph crosses the y- axis at (0 , - 14), thus c = - 14

y = x² + bx - 14

Given the graph crosses the x- axis at (2, 0), then

0 = 2² + 2b - 14

0 = 4 + 2b - 14 = 2b - 10 ( add 10 to both sides )

10 = 2b ( divide both sides by 2 )

b = 5

y = x² + 5x - 14 ← represents the graph

let y = 0 , then

x² + 5x - 14 = 0 ← in standard form

(x + 7)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 7 = 0 ⇒ x = - 7

x - 2 = 0 ⇒ x = 2

The x- intercepts are x = - 7 and x = 2

The vertex lies on the axis of symmetry which is midway between the x- intercepts, thus

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3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

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2 years ago

Which of the following symbols indicates that a value is included in a solution set.

Ulleksa [173]

Answer: Filled in bubble

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2 years ago

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