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Zina [86]
3 years ago
13

Carly is buying deli meat for her lunches this week. Turkey is on sale for $5.80 per pound. If Carly buys 1.25 pounds of turkey,

how much money does she spend?
Mathematics
2 answers:
alisha [4.7K]3 years ago
8 0

Answer:

Carly will have to spend $7.25

Step-by-step explanation:

We know that:

1 pound of turkey = $5.80

To find how much 1.25 pounds of turkey will cost, multiply:

5.80 x 1.25 = 7.25

Therefore, 1.25 pounds of turkey will cost $7.25

I really hope this helps!

-SpaceMarsh

kati45 [8]3 years ago
5 0
6.03 i believe , she gets one pound which equals to 5.80 and .25 more if you divide 5.80 with .25 you get .23 adding 5.80 with .23 equals $6.03
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The percentage difference from 239,900 to 251,895
Katarina [22]

Answer:

4.87805%

Step-by-step explanation:

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3 years ago
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Could y’all help me out with this?
Aneli [31]

Answer:

M' (5,-1)

D' (5,1)

A' (3,4)

W' (4,-1)

Step-by-step explanation:

We want to find the coordinates of Quadrilateral MDAW with vertices M(-1, 5), D(1,5), A(4,3), and W(-1,4) after a reflection across y = x.

We can find these coordinates by following the following rule:

Reflection over y = x rule : ( x , y ) ---> ( y , x )

Explanation of rule: Simply swap the places of the x and y values.

Applying rule to given coordinates:

M(-1,5) -----> swap x and y values -----> M'(5,-1)

D(1,5) ----> swap x and y values -----> D'(5,1)

A(4,3) -----> swap x and y values -----> A'(3,4)

W(-1,4) -----> swap x and y values -----> W'(4,-1)

So the coordinates of Quadrilateral MDAW  after a reflection over the y = x  line are M'(5,-1) , D'(5,1) , A'(3,4) , W'(4,-1)

For more validation, refer to the attached image

4 0
3 years ago
Given that A and B are true and X and Y are false, determine the truth value of the following proposition: ~[(A ⊃ Y) v ~(X ⊃ B)]
Anni [7]

Answer:

The value of the proposition is FALSE

Step-by-step explanation:

~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ ~X) v (B ⊃ X)]

Let's start with the smallest part: ~X. The symbol ~ is negation when X is true with the negation is false and vice-versa. In this case, ~X is true (T)

~[(A ⊃ Y) v ~(X ⊃ B)] ⋅ [~(A ≡ T) v (B ⊃ X)]

Now the parts inside parenthesis:  (A ⊃ Y),(X ⊃ B),(A ≡ T) and (B ⊃ X). The symbol ⊃ is the conditional and A ⊃ Y is false when Y is false and A is true, in any other case is true. The symbol ≡ is the biconditional and A ≡ Y is true when both A and Y are true or when both are false.

(A ⊃ Y) is False (F)

(X ⊃ B) is True (T)

(A ≡ T) is True (T)

(B ⊃ X) is False (F)

~[(F) v ~(T)] ⋅ [~(T) v (F)]

The two negations inside the brackets must be taken into account:

~[(F) v F] ⋅ [F v (F)]

The symbol left inside the brackets v is the disjunction, and A v Y is false only  with both are false. F v (F) is False.

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Again considerating the negation:

T⋅ [F]

Finally, the symbol ⋅ is the conjunction, and A v Y is true only with both are true.

T⋅ [F] is False.

5 0
3 years ago
The sum of two numbers is 156. The first number is six more
Zigmanuir [339]

Answer:

Let A equal the first number

Let B equal the second number

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B=b

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Step-by-step explanation:

Hope this helps :) Have a great day!!!!

3 0
3 years ago
Square root of 2tanxcosx-tanx=0
kobusy [5.1K]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/3242555

——————————

Solve the trigonometric equation:

\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}


Restriction for the solution:

\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.


Square both sides of  (i):

\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}

\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}


Let

\mathsf{sin\,x=t\qquad (0\le t


So the equation becomes

\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}


Solving the quadratic equation:

\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.


\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}


\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}


You can discard the negative value for  t. So the solution for  (ii)  is

\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}


Substitute back for  t = sin x.  Remember the restriction for  x:

\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}

where  k  is an integer.


I hope this helps. =)

3 0
3 years ago
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