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mr_godi [17]
2 years ago
11

4 CORIZ What are two multiples that 8 and 16 have in common?

Mathematics
1 answer:
ddd [48]2 years ago
6 0

Answer:

4 and 8

Step-by-step explanation:

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at a local green market, Debra bought 5 ounces of chinese long beans for $13.55. at that rate , how much would she pay for 8 oun
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In my opinion I think that you have to multiply 13.55 by 8 ounces to find how much she needs to pay.
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Which transformations map the strip pattern onto itself?
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 A horizontal translation and a 180 rotation.
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Let p, q, and r represent the following statements"
Nitella [24]

In logic, own symbols are used in order to be able to represent the relations between propositions in a general and independent way to the proposition, in order to be able to find the relationship process that operates in the communicated message, the propositional logic.

For this purpose there are, among others, the following logical operators: conjunction (and) ∧, disjunction (or) ∨, denial (not) ¬,  conditional (if - then) ⇒ and double conditional (if and only if, iff) ⇔.

So for this case we have:

Answer

a) Sam had pizza last night if and only if Chris finished her homework.

p⇔q

b) Pat watched the news this morning iff Sam did not have pizza last night.

r⇔¬p

c) Pat watched the news this morning if and only if Chris finished her homework and Sam did not have pizza last night.

r⇔(q∧¬p)

d) In order for Pat to watch the news this morning, it is necessary and sufficient that Sam had pizza last night and Chris finished her homework.

r⇔(p∧q)

e) <em>q ⇔ r</em>

Chris finished his homework if and only if Pat watched the news this morning

f) <em>p ⇔ (q ∧ r)</em>

Sam had pizza last night if and only if Chris finished his homework and Pat watched the news this morning

g) <em>(¬p) ⇔ (q ∨ r) </em>

Sam didn't have pizza last night if and only if Chris finished his homework or Pat watched the news this morning

h) <em>r ⇔ (p ∨ q) </em>

Pat watched the news this morning if Sam had pizza last night or Chris finished his homework

4 0
3 years ago
Ms. Hernandez gave her students a math test
emmainna [20.7K]

Answer:

85-86%

Step-by-step explanation:

Make it an equation

You want to find the percent out of a hundred, so 42 is 100 since it is all of the questions and 36 is the unknown percent X

36/42 = X/100

36             X

42             100

Cross multiply

Do math and solve for X

36 x 100 = 42 x X

3600=42X

X=85.71

7 0
2 years ago
Please help!!<br> Write a matrix representing the system of equations
frozen [14]

Answer:

(4, -1, 3)

Step-by-step explanation:

We have the system of equations:

\left\{        \begin{array}{ll}            x+2y+z =5 \\    2x-y+2z=15\\3x+y-z=8        \end{array}    \right.

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}

Now, we can multiply R₂ by -1/5. This yields:

\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}

We can now reduce R₃ by multiply it by -1/4:

\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

And finally, we can eliminate the second 1 by adding -(R₃):

\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}

Therefore, our solution set is (4, -1, 3)

And we're done!

3 0
3 years ago
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