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4 years ago

Plzzz help 4^2 × 5^2 -30

Mathematics

2 answers:

jolli1 [7]4 years ago

5 0

Answer:

370

Step-by-step explanation:

4^2 = 16

5^2 = 25

25 x 16 = 400

400 - 30 = 370

Your welcome! :D

liq [111]4 years ago

3 0

Answer:

4^2 x 5^2 - 30 = 370

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How to solve log5 x/2 = 5? i got it to 2^5=x/2 but i have NO CLUE how to solve from there?

Lerok [7]

Well
$log_a(b)=c$ translates to $a^c=b$
so
log5(x/2)=5 means 5^5=x/2, so 3125=x/2, so 6250=x

4 0

3 years ago

20 feet of ribbon, cutting ribbon into 7 1/2 inches or 6 3/4 inch lengths to make bracelets. Write an algebraic expression that

IRINA_888 [86]

<u>ANSWER: </u>

The algebraic expression for number of ribbons is $\frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}$ and the algebraic expression for length of ribbon is $\frac{25 x}{3} \text { feet }$

<u>SOLUTION: </u>

Given, 20 feet of ribbon, cutting ribbon into $7\frac{1}{2}$ inches or $6\frac{3}{4}$ inch lengths to make bracelets.

Let us convert the mixed fraction to improper fractions.

$7 \frac{1}{2}=\frac{7 \times 2+1}{2}=\frac{15}{2} \text { and } 6 \frac{3}{4}=\frac{6 \times 4+3}{4}=\frac{27}{4}$

Let, the length of bracelets can be made be “x”

First we have to remove the excess length of ribbon and then we have to divide the remaining with length of bracelet we want.

$\text { number of ribbons }=\frac{\text { length of ribbon-excess ribbon.}}{\text {length of bracelet}}$

$\left.=\frac{20 \text { feet }-\left\{\frac{20 \text { feep }}{x}\right.}{x} \text { [we know that, }\{x\} \text { is fractional part of } x\right$ ]

So, the algebraic expression for number of ribbons is $\frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}$

Now, let us find the algebraic expression for feet of ribbon to make 100 ribbons.

We have 100 bracelets of length x, then total length = 100 × x

length of ribbon = 100x inches

$\begin{array}{l}{=100 \mathrm{x} \times \frac{1}{12} \text { feet }} \\\\ {=\frac{25 x}{3} \text { feet }}\end{array}$

So, the algebraic expression for length of ribbon is $\frac{25 x}{3} \text { feet }$

Hence, the algebraic expression for number of ribbons is $\frac{20 \text { feet }-\left\{\frac{20 \text { feet }}{x}\right\}}{x}$ and the algebraic expression for length of ribbon is $\frac{25 x}{3} \text { feet }$

5 0

3 years ago

Solve the equations 11(x+10)=132

VladimirAG [237]

11(x+10) = 132

Distributive property

11x + 110 = 132

Now isolate the variable.

11x + 110 = 132
-110 -110

132 - 110 = 22

11x = 22

11x/11 = 22/11
x= 2

Answer: x = 2

4 0

4 years ago

Read 2 more answers

A lifeguard is in a look out chair and sees a person in distress. The eye level of the lifeguard is 15 feet above the ground. Th

faltersainse [42]

Answer:

Horizontal distance between lifeguard and the person is 22 feet.

Step-by-step explanation:

Given: A lifeguard sees a person in distress.The eye level of the lifeguard is 15 feet above the ground. Angle of depression is 34°.

To find: Horizontal distance between lifeguard and the person.

Solution : If we draw a triangle then tan∅ = $\frac{height}{base}$

Here base is the horizontal distance.

Now we put the values in the formula.

tan 34° = $\frac{15}{base}$

Or Base = $\frac{15}{tan34}$

= $\frac{15}{0.675} = 22.22 feet$

So the answer is 22 feet.

7 0

3 years ago

Let v⃗ 1=⎡⎣⎢033⎤⎦⎥,v⃗ 2=⎡⎣⎢1−10⎤⎦⎥,v⃗ 3=⎡⎣⎢30−3⎤⎦⎥ be eigenvectors of the matrix A which correspond to the eigenvalues λ1=−1, λ2

kaheart [24]

Answer:

- x as a linear combination :

x = -1 v1+ 0 v2+ 1 v3.

- Transpose Ax = (12, -6, -6)

Step-by-step explanation:

Given v1 = (0 3 3),v2 = (1 −1 0), v3 = (3 0 −3) be eigenvectors of the matrix A which correspond to the eigenvalues λ1 = −1, λ2 = 0, and λ3 = 1, respectively, and let x = (−2 −4 0). Express x as a linear combination of v1, v2, and v3, and find Ax .

To write x as a linear combination of v1, v2, and v3

x = -1 v1+ 0 v2+ 1 v3.

To find Ax

Write A = (0 ......3 ......3 )

...................(1 ......-1 ......0)

...................(3 ......0......-3)

Since transpose x = (-2, 4, 0)

Ax =......... (0 ......3......3 )(-2)

...................(1 ......-1 ......0)(4)

...................(3 ......0......-3)(0)

= (0×-2 + 3×4 + 3×0)

...(1×-2 + -1×4 + 0×0)

.. (3×-2 + 0×4 + -3×0)

As = (12)

....(-6)

Transpose Ax = (12, -6, -6)

7 0

3 years ago

Plzzz help 4^2 × 5^2 -30​

Plzzz help 4^2 × 5^2 -30