Ask question

Ask question

All categories

2 years ago

13

your teacher is giving you a test with 100 points containing 40 questions. there are two point and four point questions on the t

est. how many of each type of question on the test?

1 answer:

tino4ka555 [31]2 years ago

5 0

Answer:

30 of 2 points= 60 points

10 of 4 points = 40points so its 30 of 2 points and 10 of 4

Step-by-step explanation:

You might be interested in

What is the slope-intercept form of the equation of a line that passes through (5, -4) and has a slope of 34?

Alla [95]

One way to find the slope-intercept form is to plug the given values into point-slope form, y - y_{1} = m (x - x_{1}), where x_{1} and y_{1} are the coordinate points and m is the slope. Then, you solve for y.

y - y_{1} = m (x - x_{1}) Plug in the values
y - (-4) = 34 (x - 5) Fix the minus negative four
y + 4 = 34 (x - 5) Use the Distributive Property
y + 4 = 34x - 170 Subtract 4 from both sides
y = 34x - 174

The slope-intercept form of the equation that passes through (5, -4) and has a slope of 34 is y = 34x - 174.

6 0

2 years ago

xeze [42]

Answer:

54

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

What is the slope of this line?<br><br> A. 1/3<br> B. 1/2<br> C. 1<br> D. 2

HACTEHA [7]

Slope = rise / run

so answer is

C. 1

6 0

2 years ago

Read 2 more answers

Plz show work and solve quick please need help!!!!!

Simora [160]

Surface area is just the area of all these 4 triangles plus the rectangle.

First we can find the area of the rectangle.

$\sf A=lw$

Half of the length is 28 cm, so the full length must be 28 * 2 = 56 cm.

$\sf A=(56)(27)$

$\sf A=1512cm^2$

The base for the left and right triangles are 27. The heights would be the net length minus half the length of the rectangle:

$\sf 82.8-28=54.8\? cm$

Calculate the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(27)(54.8)$

$\sf A=739.8\?cm^2$

We have two of these triangles.

$\sf 739.8\times 2=1479.6\? cm^2$

Now do the other two pair of triangles. The bases for them are 28 + 28 = 56 cm. The heights would be the net width minus the width of the rectangle:

$\sf 81.2-27=54.2\?cm$

Now find the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(56)(54.2)$

$\sf A=1517.6\? cm^2$

We have two of these triangles.

$\sf 1517.6\times 2=3035.2\?cm^2$

Add all the areas together:

$\sf 1512+1479.6+3035.2=\boxed{\sf 6026.80\? cm^2}$

7 0

3 years ago

I need help :)<br> Find the GCF: 10x^5−16x^4+4x^2

andreev551 [17]

The greatest common factor is 2x^2.
the largest number that divides evenly into <span>10x^5−16x^4+4x^2 is two.
the highest degree is x^2 so the answer would be 2x^2
</span>

5 0

2 years ago