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skad [1K]
3 years ago
10

mrs. Burns is making cookies. The recipe calls for three 1/2 cups of flour and three 1/4 cup of sugar. if she wants to double th

e recipe, How much flour and sugar does Miss Burns need?how much flour and sugar does Mrs Burns need?
Mathematics
2 answers:
sp2606 [1]3 years ago
6 0
1 cup of flour and 1/2 cup of sugar
Maslowich3 years ago
5 0

Mrs. Burns needs 3 cups of flour and 1 1/2 cups of sugar.

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3 determine the highest real root of f (x) = x3− 6x2 + 11x − 6.1: (a) graphically. (b) using the newton-raphson method (three it
Juliette [100K]

(a) See the first attachment for a graph. This graphing calculator displays roots to 3 decimal places. (The third attachment shows a different graphing calculator and 10 significant digits.)

(b) In the table of the first attachment, the column headed by g(x) gives iterations of Newton's Method. (For Newton's method, it is convenient to let the calculator's derivative function compute the derivative f'(x) of the function f(x). We have defined g(x) = x - f(x)/f'(x).) The result of the 3rd iteration is ...

... x ≈ 3.0473167

(c) The function h(x₁, x₂) computes iterations using the secant method. The results for three iterations of that method are shown below the table in the attachment. The result of the 3rd iteration is ...

... x ≈ 3.2291234

(d) The function h(x, x+0.01) computes the modified secant method as required by the problem statement. The result of the 3rd iteration is ...

... x ≈ 3.0477377

(e) Using <em>Mathematica</em>, the roots are found to be as shown in the second attachment. The highest root is about ...

... x ≈ 3.0466805180

_____

<em>Comment on these methods</em>

Newton's method can have convergence problems if the starting point is not sufficiently close to the root. A graphing calculator that gives a 3-digit approximation (or better) can help avoid this issue. For the calculator used here, the output of "g(x)" is computed even as the input is typed, so one can simply copy the function output to the input to get a 12-significant digit approximation of the root as fast as you can type it.

The "modified" secant method is a variation of the secant method that does not require two values of the function to start with. Instead, it uses a value of x that is "close" to the one given. For our purpose here, we can use the same h(x1, x2) for both methods, with a different x2 for the modified method.

We have defined h(x1, x2) = x1 - f(x1)(f(x1)-f(x2))/(x1 -x2).

6 0
2 years ago
Harry took out a loan from the bank. the variable ddd models harry's remaining debt (in dollars) ttt months after he took out th
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Answer:

Harry has a loan of $9000 in total. Harry obtained a loan from the bank. Explanation Harry's remaining debt, expressed in dollars, is modeled as a function of time t, expressed in months, by the function D(t). The role is played by, This function can be used to determine that $200 is being subtracted each month from the function, meaning Harry is paying $200 toward his loan. Harry has not yet made any payments, therefore we may set t=0 to obtain the total amount of his solo. Therefore, the value of D(t) will reveal the loan's net amount. Harry's borrowing, therefore, equals to $9000.

5 0
2 years ago
Evaluate the expression<br> 7+10√1/25
nata0808 [166]
The answer is 9

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4 0
2 years ago
Find each measure! Please please help me!!! 40 points!! No files!!!
ANTONII [103]

Answer:

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Step-by-step explanation:

If you want i can explain in the comments.

5 0
2 years ago
State whether the triangles could be proven congruent, if possible, by SSS or SAS.
Helen [10]

Answer:

1. SSS

2. SAS

3. SAS

4. SSS

5. SAS

6. SAS

Step-by-step explanation:

1) Side FY ≅ Side CW  Given

Side FP ≅ Side CM     Given

Side YP ≅ Side MW    Given

∴ΔMCW ≅ ΔFPY by the Side-Side-Side (SSS) rule of congruency

2) ∠CBD ≅ ∠BCA given that both are alternate interior angles

Side EB ≅ Side EC and Side DB ≅ Side CA Given

ΔBED ≅ ΔAEC by Side-Angle-Side (SAS) rule of congruency

3. ∠SVU ≅ ∠SVT   Given

Side SV ≅ Side SV  by reflexive property

Side VT ≅ Side VU   Given

∴ ΔVSU ≅ ΔVST by Side-Angle-Side (SAS) rule of congruency

4. Side MN ≅ Side QP  Given

Side MQ ≅ Side NP     Given

Side NQ ≅ Side NQ    by reflexive property

∴ΔQNM ≅ ΔQNP by the Side-Side-Side (SSS) rule of congruency

5. Indirect proof

Side GL ≅ Side HL  Given

Side GJ ≅ Side HK     Given

∠JLG ≅ ∠HLK    vertically opposite angles

By sine rule

GJ/sin(∠JLG) = GL/n(∠GJL)

Similarly

HK/sin(∠HLK) = HL/n(∠HKL)

∴ ∠GJL ≅ ∠HKL

∴ ∠LGJ ≅ ∠LHK third angle of two triangles given the other two angles are congruent

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6. ∠XZY ≅ ∠XZW   Supplementary ∠s with ∠XZY = 90°

Side XZ ≅ Side XZ  by reflexive property

Side ZW ≅ Side ZY   Given

∴ ΔXYZ ≅ ΔXWZ by Side-Angle-Side (SAS) rule of congruency.

3 0
2 years ago
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