4. Jan bought donuts for $6.75. Each bagel cost $0.75. How many bagels did she buy?

Mathematics

1 answer:

Nataly_w [17]3 years ago

6 0

Answer:

9 bagels

Step-by-step explanation:

Divide the price ($6.75) by the price of one bagel ($0.75) to get the amount of bagels Jan bought. (9)

You might be interested in

How do you solve all of these ?

Flura [38]

5)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 5}}\quad ,&{{ -3}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies -1
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-3)=-1(x-5)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y+3=-x+5\implies y=-x+2$

6)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ 1}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 3
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-1=3(x-2)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-1=3x-6\implies y=3x-5$

7)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 5}}\quad ,&{{ 2}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 0
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-2=0(x-5)
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-2=0\implies y=2$

8)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -2}}\quad ,&{{ 0}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-0=\cfrac{5}{2}(x-(-2))
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=\cfrac{5}{2}(x+2)\implies y=\cfrac{5}{2}x+5$

9)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ -2}}\quad ,&{{ 0}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-0=\cfrac{5}{2}(x-(-2))
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y=\cfrac{5}{2}(x+2)\implies y=\cfrac{5}{2}x+5$

10)

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{-2}}\quad ,&{{ -2}})\quad 
% (c,d)
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{3}{2}
\\\\\\
% point-slope intercept
y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-(-2)=\cfrac{3}{2}(x-(-2))
\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y+2=\cfrac{3}{2}(x+2)\implies y+2=\cfrac{3}{2}x+3\implies y=\cfrac{3}{2}x+1$

4 0

3 years ago

What is the exponent in the expression 7^6 ? 6 7 13 42

Illusion [34]

The exponent is 6, since 7^6 is the same as (see picture).

5 0

3 years ago

Read 2 more answers

The additive inverse of the complex number a+bi is -a-bi. The sum of a complex number and its additive inverse is zero.

atroni [7]

I think it is 2-8i
Hope this helps

8 0

3 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

MariettaO [177]

The given equation for the relationship between a planet's orbital period, T and the planet's mean distance from the sun, A is T^2 = A^3. Let the orbital period of planet X be T(X) and that of planet Y = T(Y) and let the mean distance of planet X from the sun be A(X) and that of planet Y = A(Y), then A(Y) = 2A(X) [T(Y)]^2 = [A(Y)]^3 = [2A(X)]^3 But [T(X)]^2 = [A(X)]^3 Thus [T(Y)]^2 = 2^3[T(X)]^2 [T(Y)]^2 / [T(X)]^2 = 2^3 T(Y) / T(X) = 2^3/2 Therefore, the orbital period increased by a factor of 2^3/2 <span>
</span>

6 0

3 years ago

Read 2 more answers

How many dots are in the 30th figure

mafiozo [28]

Hey can you show a picture of the figure?

4 0

3 years ago